Problem:
Single Source Shortest Path II For a given weighted graph $G = (V, E)$, find the shortest path from a source to each vertex. For each vertex $u$, print the total weight of edges on the shortest path from vertex $0$ to $u$. Input In the first line, an integer $n$ denoting the number of vertices in $G$ is given. In the following $n$ lines, adjacency lists for each vertex $u$ are respectively given in the following format: $u$ $k$ $v_1$ $c_1$ $v_2$ $c_2$ ... $v_k$ $c_k$ Vertices in $G$ are named with IDs $0, 1, ..., n-1$. $u$ is ID of the target vertex and $k$ denotes its degree. $v_i (i = 1, 2, ... k)$ denote IDs of vertices adjacent to $u$ and $c_i$ denotes the weight of a directed edge connecting $u$ and $v_i$ (from $u$ to $v_i$). Output For each vertex, print its ID and the distance separated by a space character in a line respectively. Print in order of vertex IDs. Constraints  $1 \leq n \leq 10,000$  $0 \leq c_i \leq 100,000$  $|E| < 500,000$  All vertices are reachable from vertex $0$  Sample Input 1 5
0 3 2 3 3 1 1 2
1 2 0 2 3 4
2 3 0 3 3 1 4 1
3 4 2 1 0 1 1 4 4 3
4 2 2 1 3 3 Sample Output 1 0 0
1 2
2 2
3 1
4 3 Reference Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution (Python):

n = int(input())
inf = 1000000000000000

def minHeapify(Q,i):
    H = len(Q)
    l = 2*i
    r = 2*i + 1
    if l < H and Q[l][1] < Q[i][1]:
        smallest = l
    else:
        smallest = i
    if r < H and Q[r][1] < Q[smallest][1]:
        smallest = r

    if smallest != i:
        Q[i], Q[smallest] = Q[smallest], Q[i]
        minHeapify(Q,smallest)

def buildMinHeap(Q):
    H = len(Q)
    for i in range(H // 2, -1, -1):
        minHeapify(Q,i)

def Insert(Q,key):
    v = key[0]
    Q.append([v,inf])
    lens = len(Q)
    heapIncreaseKey(Q,lens-1,key)

def heapIncreaseKey(Q,i,key):
    Q[i][1] = key[1]
    while i > 0 and Q[i // 2][1] > Q[i][1]:
        Q[i], Q[i // 2] = Q[i // 2], Q[i]
        i //= 2

def heapExtractMin(Q):
    H = len(Q)
    minv = Q[0][0]
    Q[0] = Q[H-1]
    Q.pop()
    minHeapify(Q,0)
    return minv




def dijkstra(s):
    color = ["white" for i in range(n)]
    d = [inf for i in range(n)]
    d[s] = 0
    Q = [[s,0]]
    while Q != []:
        u = heapExtractMin(Q)
        color[u] = "black"
        for vw in mtrx[u]:
            [v,w] = vw
            if color[v] != "black":
                if d[u] + w < d[v]:
                    d[v] = d[u] + w
                    color[v] = "gray"
                    Insert(Q,[v,d[v]])
    return d



mtrx = [[] for i in range(n)]
for i in range(n):
    col = list(map(int,input().split()))
    u,k = col[0], col[1]
    for j in range(k):
        [v,w] = col[2*j+2: 2*j+4]
        mtrx[u].append([v,w])



d = dijkstra(0)
for i,j in enumerate(d):
    print(i,j)

Solution (Python):


import heapq

def dj(adjm, start):
    n = len(adjm)
    d = [float("inf")] * n
    d[start] = 0
    q = []
    heapq.heappush(q, (0, start))

    while(q):
        td, v = heapq.heappop(q)
        for k in adjm[v].keys():
            nd = td + adjm[v][k]
            if d[k] > nd:
                d[k] = nd
                heapq.heappush(q, (nd, k))
    return d


def main():
    n = int(input())
    
    adjm = [{} for _ in range(n)]
    for i in range(n):
        vdata = list(map(int, input().split()))
        for j in range(vdata[1]):
            v, c = vdata[j * 2 + 2], vdata[j * 2 + 3]
            adjm[i][v] = c
    ans = dj(adjm, 0)
    for i, d in enumerate(ans):
        print(i, d)


if __name__ == "__main__":
    main()

Solution (Python):

from heapq import heappop, heappush


def dijkstra(edges, start, *, adj_matrix=False, default_value=float('inf')):
    '''
    Returns best costs to each node from 'start' node in the given graph.
    (Single Source Shortest Path - SSSP)

    If edges is given as an adjacency list including costs and destination nodes on possible edges, adj_matrix should be False(default), and it is generally when this functions works much faster.

    Note that costs in edges should follow the rules below
    1. The cost from a node itself should be 0,
    2. If there is no edge between nodes, the cost between them should be default_value
    3. The costs can't be negative

    
    edges = [[(1, 2), (2, 5), (3, 4)],         
             [(0, 2), (3, 3), (4, 6)],         
             [(0, 5), (3, 2), (5, 6)],         
             [(0, 4), (1, 3), (2, 2), (4, 2)], 
             [(1, 6), (3, 2), (5, 4)],         
             [(2, 6), (4, 4)]]                 

    
    edges = [[0, 2, 5, 4, inf, inf],           
             [2, 0, inf, 3, 6, inf],           
             [5, inf, 0, 2, inf, 6],           
             [4, 3, 2, 0, 2, inf],             
             [inf, 6, inf, 2, 0, 4],           
             [inf, inf, 6, inf, 4, 0]]         
    '''

    n = len(edges)

    inf = float('inf')
    costs = [inf] * n
    costs[start] = 0
    pq, rem = [(0, start)], n - 1

    while pq and rem:
        tmp_cost, tmp_node = heappop(pq)
        if costs[tmp_node] < tmp_cost:
            continue
        rem -= 1

        nxt_edges = ((node, cost) for node, cost in enumerate(edges[tmp_node]) if cost != default_value) if adj_matrix else\
                    edges[tmp_node]

        for nxt_node, nxt_cost in nxt_edges:
            new_cost = tmp_cost + nxt_cost
            if costs[nxt_node] > new_cost:
                costs[nxt_node] = new_cost
                heappush(pq, (new_cost, nxt_node))

    return costs


def dijkstra_route(edges,
                   start,
                   goal,
                   *,
                   adj_matrix=False,
                   default_value=float('inf'),
                   verbose=False):
    '''
    Trys to find the best route to the 'goal' from the 'start'
    '''

    n = len(edges)

    inf = float('inf')
    costs = [inf] * n
    costs[start] = 0
    pq, rem = [(0, start)], n - 1
    prevs = [-1 for _ in [None] * n]

    while pq and rem:
        tmp_cost, tmp_node = heappop(pq)
        if costs[tmp_node] < tmp_cost:
            continue
        rem -= 1

        nxt_edges = ((node, cost) for node, cost in enumerate(edges[tmp_node]) if cost != default_value) if adj_matrix else\
                    edges[tmp_node]

        for nxt_node, nxt_cost in nxt_edges:
            new_cost = tmp_cost + nxt_cost
            if costs[nxt_node] > new_cost:
                costs[nxt_node] = new_cost
                heappush(pq, (new_cost, nxt_node))
                prevs[nxt_node] = tmp_node

    min_route = []
    prev = goal
    cnt = 0
    while prev != start:
        min_route.append(prev)
        prev = prevs[prev]

        cnt += 1
        if prev == -1 or cnt > n:
            raise NoRouteError(
                'There is no possible route in this graph \nedges: {} \nstart: {} \ngoal: {}'.
                format(edges, start, goal))

    else:
        min_route.append(prev)
    min_route = min_route[::-1]
    min_cost = costs[goal]

    if verbose:
        print('---route---')
        for node in min_route[:-1]:
            print('{} -> '.format(node), end='')
        else:
            print(min_route[-1])

        print('---distance---')
        print(min_cost)

    return costs, min_route


class NoRouteError(Exception):
    pass


if __name__ == '__main__':
    '''
    an example of using this function
    AIZU ONLINE JUDGE - ALDS_1_12_C
    '''

    n = int(input())
    edges = [[] for i in range(n)]
    for i in range(n):
        i, k, *kedges = map(int, input().split())
        for edge in zip(kedges[::2], kedges[1::2]):
            edges[i].append(edge)

    for i, cost in enumerate(dijkstra(edges, 0)):
        print(i, cost)

Solution (Python):

"""
Created on Sat Jul 14 19:11:44 2018

@author: maezawa
"""
infinity = 2*10**9

n = int(input())
adj = [[] for _ in range(n)]

for j in range(n):
    ain = list(map(int, input().split()))
    u = ain[0]
    k = ain[1]
    for i in range(2,2+2*k,2):
        adj[u].append([ain[i],ain[i+1]])
    

color = [0]*n
d = [infinity for _ in range(n)]
p = [0]*n
p[0] = -1
d[0] = 0

def dyk(s):
    cnt = 1
    cur = s
    candidates = set() 
    while True:
        for i, w in adj[cur]:
            if color[i] == 2:
                continue
            if d[i] > d[cur] + w:
                d[i] = d[cur] + w
                color[i] = 1
            candidates.add(i)
        min_d = infinity
        for j in candidates:
            if min_d > d[j]:
                min_node = j
                min_d = d[j]
        candidates.remove(min_node)
        p[min_node] = cur
        color[cur] = 2
        cur = min_node
        cnt += 1
        
        
        if cnt == n:
            break
    return

dyk(0)
for i in range(n):
    print(i, d[i])

Solution (Python):

from collections import defaultdict
from heapq import heappop, heappush


def dijkstra(edges, start, *, adj_matrix=False, default_value=float('inf')):
    '''
    Returns best costs to each node from 'start' node in the given graph.
    (Single Source Shortest Path - SSSP)

    If edges is given as an adjacency list including costs and destination nodes on possible edges, adj_matrix should be False(default), and it is generally when this functions works much faster.

    Note that costs in edges should follow the rules below
    1. The cost from a node itself should be 0,
    2. If there is no edge between nodes, the cost between them should be default_value
    3. The costs can't be negative

    
    edges = [[(1, 2), (2, 5), (3, 4)],         
             [(0, 2), (3, 3), (4, 6)],         
             [(0, 5), (3, 2), (5, 6)],         
             [(0, 4), (1, 3), (2, 2), (4, 2)], 
             [(1, 6), (3, 2), (5, 4)],         
             [(2, 6), (4, 4)]]                 

    
    edges = [[0, 2, 5, 4, inf, inf],           
             [2, 0, inf, 3, 6, inf],           
             [5, inf, 0, 2, inf, 6],           
             [4, 3, 2, 0, 2, inf],             
             [inf, 6, inf, 2, 0, 4],           
             [inf, inf, 6, inf, 4, 0]]         
    '''

    n = len(edges)

    inf = float('inf')
    
    costs = defaultdict(lambda: inf)
    costs[start] = 0
    pq, rem = [(0, start)], n - 1

    while pq and rem:
        tmp_cost, tmp_node = heappop(pq)
        if costs[tmp_node] < tmp_cost:
            continue
        rem -= 1

        nxt_edges = ((node, cost) for node, cost in enumerate(edges[tmp_node]) if cost != default_value) if adj_matrix else\
                    edges[tmp_node]

        for nxt_node, nxt_cost in nxt_edges:
            new_cost = tmp_cost + nxt_cost
            if costs[nxt_node] > new_cost:
                costs[nxt_node] = new_cost
                heappush(pq, (new_cost, nxt_node))

    return costs


def dijkstra_route(edges,
                   start,
                   goal,
                   *,
                   adj_matrix=False,
                   default_value=float('inf'),
                   verbose=False):
    '''
    Trys to find the best route to the 'goal' from the 'start'
    '''

    n = len(edges)

    inf = float('inf')
    
    costs = defaultdict(lambda: inf)
    costs[start] = 0
    pq, rem = [(0, start)], n - 1
    
    prevs = defaultdict(lambda: -1)

    while pq and rem:
        tmp_cost, tmp_node = heappop(pq)
        if costs[tmp_node] < tmp_cost:
            continue
        rem -= 1

        nxt_edges = ((node, cost) for node, cost in enumerate(edges[tmp_node]) if cost != default_value) if adj_matrix else\
                    edges[tmp_node]

        for nxt_node, nxt_cost in nxt_edges:
            new_cost = tmp_cost + nxt_cost
            if costs[nxt_node] > new_cost:
                costs[nxt_node] = new_cost
                heappush(pq, (new_cost, nxt_node))
                prevs[nxt_node] = tmp_node

    min_route = []
    prev = goal
    cnt = 0
    while prev != start:
        min_route.append(prev)
        prev = prevs[prev]

        cnt += 1
        if prev == -1 or cnt > n:
            raise NoRouteError(
                'There is no possible route in this graph \nedges: {} \nstart: {} \ngoal: {}'.
                format(edges, start, goal))

    else:
        min_route.append(prev)
    min_route = min_route[::-1]
    min_cost = costs[goal]

    if verbose:
        print('---route---')
        for node in min_route[:-1]:
            print('{} -> '.format(node), end='')
        else:
            print(min_route[-1])

        print('---distance---')
        print(min_cost)

    return costs, min_route


class NoRouteError(Exception):
    pass


if __name__ == '__main__':
    '''
    an example of using this function
    AIZU ONLINE JUDGE - ALDS_1_12_C
    '''

    n = int(input())
    edges = [[] for i in range(n)]
    for i in range(n):
        i, k, *kedges = map(int, input().split())
        for edge in zip(kedges[::2], kedges[1::2]):
            edges[i].append(edge)

    
    
    for i, cost in sorted(dijkstra(edges, 0).items(), key=lambda x: x[0]):
        print(i, cost)

Solution (Python):

from heapq import heappop, heappush


def dijkstra(edges, start, *, adj_matrix=False, default_value=float('inf')):
    '''
    Returns best costs to each node from 'start' node in the given graph.
    (Single Source Shortest Path - SSSP)

    If edges is given as an adjacency list including costs and destination nodes on possible edges, adj_matrix should be False(default), and it is generally when this functions works much faster.

    Note that costs in edges should follow the rules below
    1. The cost from a node itself should be 0,
    2. If there is no edge between nodes, the cost between them should be default_value
    3. The costs can't be negative

    
    edges = [[(1, 2), (2, 5), (3, 4)],         
             [(0, 2), (3, 3), (4, 6)],         
             [(0, 5), (3, 2), (5, 6)],         
             [(0, 4), (1, 3), (2, 2), (4, 2)], 
             [(1, 6), (3, 2), (5, 4)],         
             [(2, 6), (4, 4)]]                 

    
    edges = [[0, 2, 5, 4, inf, inf],           
             [2, 0, inf, 3, 6, inf],           
             [5, inf, 0, 2, inf, 6],           
             [4, 3, 2, 0, 2, inf],             
             [inf, 6, inf, 2, 0, 4],           
             [inf, inf, 6, inf, 4, 0]]         
    '''

    n = len(edges)

    costs = [float('inf')] * n
    costs[start] = 0

    pq = [(0, start)]
    rem = n - 1

    while pq and rem:
        tmp_cost, tmp_node = heappop(pq)
        if costs[tmp_node] < tmp_cost:
            continue
        rem -= 1

        nxt_edges = ((node, cost) for node, cost in enumerate(edges[tmp_node]) if cost != default_value) if adj_matrix else\
                    edges[tmp_node]

        for nxt_node, nxt_cost in nxt_edges:
            new_cost = tmp_cost + nxt_cost
            if costs[nxt_node] > new_cost:
                costs[nxt_node] = new_cost
                heappush(pq, (new_cost, nxt_node))

    return costs


def dijkstra_route(edges,
                   start,
                   goal,
                   *,
                   adj_matrix=False,
                   default_value=float('inf'),
                   verbose=False):
    '''
    Trys to find the best route to the 'goal' from the 'start'
    '''

    n = len(edges)

    costs = [float('inf')] * n
    costs[start] = 0
    pq = [(0, start)]
    rem = n - 1
    prevs = [-1 for _ in [None] * n]

    while pq and rem:
        tmp_cost, tmp_node = heappop(pq)
        if costs[tmp_node] < tmp_cost:
            continue
        rem -= 1

        nxt_edges = ((node, cost) for node, cost in enumerate(edges[tmp_node]) if cost != default_value) if adj_matrix else\
                    edges[tmp_node]

        for nxt_node, nxt_cost in nxt_edges:
            new_cost = tmp_cost + nxt_cost
            if costs[nxt_node] > new_cost:
                costs[nxt_node] = new_cost
                heappush(pq, (new_cost, nxt_node))
                prevs[nxt_node] = tmp_node

    min_route = []
    prev = goal
    cnt = 0
    while prev != start:
        min_route.append(prev)
        prev = prevs[prev]

        cnt += 1
        if prev == -1 or cnt > n:
            raise NoRouteError(
                'There is no possible route in this graph \nedges: {} \nstart: {} \ngoal: {}'.
                format(edges, start, goal))

    else:
        min_route.append(prev)
    min_route = min_route[::-1]
    min_cost = costs[goal]

    if verbose:
        print('---route---')
        for node in min_route[:-1]:
            print('{} -> '.format(node), end='')
        else:
            print(min_route[-1])

        print('---distance---')
        print(min_cost)

    return costs, min_route


class NoRouteError(Exception):
    pass


if __name__ == '__main__':
    '''
    an example of using this function
    AIZU ONLINE JUDGE - ALDS_1_12_C
    '''

    n = int(input())
    edges = [[] for i in range(n)]
    for i in range(n):
        i, k, *kedges = map(int, input().split())
        for edge in zip(kedges[::2], kedges[1::2]):
            edges[i].append(edge)

    costs = dijkstra(edges, 0)
    for i, cost in enumerate(costs):
        print(i, cost)

Solution (Python):
import sys

def dijkstra(v_count: int, edges: list, start: int,
             *, adj_matrix: bool = False, unreachable=float("inf")) -> list:
    """ ダイクストラ法
    :param v_count: 頂点数
    :param edges: 辺のリスト(隣接リストor隣接行列)
    :param start: スタートする頂点
    :param adj_matrix: edgesに渡したリストが隣接行列ならTrue
    :param unreachable: 到達不能を表すコスト値
                        隣接行列の辺の有無の判定および返すリストの初期値に使用
    """
    from heapq import heappush, heappop

    vertices = [unreachable] * v_count
    vertices[start] = 0
    q, rem = [(0, start)], v_count - 1

    while q and rem:
        cost, v = heappop(q)
        if vertices[v] < cost:
            continue
        rem -= 1

        dests = (filter(lambda x: x[1] != unreachable, enumerate(edges[v]))
                 if adj_matrix else edges[v])

        for dest, _cost in dests:
            newcost = cost + _cost
            if vertices[dest] > newcost:
                vertices[dest] = newcost
                heappush(q, (newcost, dest))

    return vertices

n = int(input())
edges = [[] for _ in [0]*n]
for a in (tuple(map(int, l.split())) for l in sys.stdin):
    edges[a[0]] = tuple(zip(a[2::2], a[3::2]))
vertices = dijkstra(n, edges, 0)
for i, n in enumerate(vertices):
    print(i, n)

Solution (Python):
import sys
import heapq
n = int(input())
V = [[int(i) for i in input().split()] for _ in range(n)]
dis = [sys.maxsize for _ in range(n)]
dis[0] = 0
h = []
for i in range(n):
    heapq.heappush(h,(dis[i],i))
while(len(h)>0):
    node = heapq.heappop(h)
    tmp = node[1]
    for i in range(V[tmp][1]):
        ind = V[tmp][2*i+2]
        cos = V[tmp][2*i+3]
        if cos + dis[tmp] < dis[ind]:
            dis[ind] = cos + dis[tmp]
            heapq.heappush(h,(dis[ind],ind))
for i,j in enumerate(dis):
    print(i,j)

Solution (Python):
from sys import maxsize
from heapq import heappop, heappush


class Node:
    WHITE = 0
    GRAY = 1
    BLACK = 2

    def __init__(self):
        self.adjs = []
        self.color = Node.WHITE
        self.dist = maxsize

    def __lt__(self, other):
        return self.dist < other.dist

    def __repr__(self):
        return "Node({},{},{})".format(self.adjs, self.color, self.dist)


def read_adj():
    n = int(input())
    nodes = [Node() for _ in range(n)]

    for i in range(n):
        u, _, *rest = [int(x) for x in input().split()]
        for i in range(0, len(rest) - 1, 2):
            nodes[u].adjs.append((nodes[rest[i]], rest[i+1]))

    return nodes


def shortest_path(nodes):
    h = []

    nodes[0].dist = 0
    nodes[0].color = Node.GRAY
    heappush(h, nodes[0])

    while True:
        while len(h) > 0:
            min_node = heappop(h)
            if min_node.color != Node.BLACK:
                break
        else:
            break

        min_node.color = Node.BLACK

        for adj, c in min_node.adjs:
            if min_node.dist + c < adj.dist:
                adj.dist = min_node.dist + c
                adj.color = Node.GRAY
                heappush(h, adj)


def print_dist(nodes):
    for u, n in enumerate(nodes):
        print(u, n.dist)


def main():
    nodes = read_adj()
    shortest_path(nodes)
    print_dist(nodes)


if __name__ == '__main__':
    main()

Solution (Python):
from sys import maxsize
from heapq import heappop, heappush


class Node:
    WHITE = 0
    GRAY = 1
    BLACK = 2

    def __init__(self):
        self.adjs = []
        self.color = Node.WHITE
        self.dist = maxsize

    def __lt__(self, other):
        return self.dist < other.dist

    def __repr__(self):
        return "Node({},{},{})".format(self.adjs, self.color, self.dist)


def read_adj():
    n = int(input())
    nodes = [Node() for _ in range(n)]

    for i in range(n):
        u, _, *rest = [int(x) for x in input().split()]
        for i in range(0, len(rest) - 1, 2):
            nodes[u].adjs.append((rest[i], rest[i+1]))

    return nodes


def shortest_path(nodes):
    h = []

    nodes[0].dist = 0
    nodes[0].color = Node.GRAY
    heappush(h, (0, 0))

    while len(h) > 0:
        min_node, min_dist = heappop(h)
        if nodes[min_node].dist < min_dist:
            continue

        nodes[min_node].color = Node.BLACK

        for adj, c in nodes[min_node].adjs:
            if nodes[min_node].dist + c < nodes[adj].dist:
                nodes[adj].dist = nodes[min_node].dist + c
                nodes[adj].color = Node.GRAY
                heappush(h, (adj, nodes[adj].dist))


def print_dist(nodes):
    for u, n in enumerate(nodes):
        print(u, n.dist)


def main():
    nodes = read_adj()
    shortest_path(nodes)
    print_dist(nodes)


if __name__ == '__main__':
    main()

Solution (Python):
from sys import maxsize
from heapq import heappop, heappush


class Node:
    WHITE = 0
    GRAY = 1
    BLACK = 2

    def __init__(self):
        self.adjs = []
        self.color = Node.WHITE
        self.dist = maxsize

    def __lt__(self, other):
        return self.dist < other.dist

    def __repr__(self):
        return "Node({},{},{})".format(self.adjs, self.color, self.dist)


def read_adj():
    n = int(input())
    nodes = [Node() for _ in range(n)]

    for i in range(n):
        u, _, *rest = [int(x) for x in input().split()]
        for i in range(0, len(rest) - 1, 2):
            nodes[u].adjs.append((rest[i], rest[i+1]))

    return nodes


def shortest_path(nodes):
    h = []

    nodes[0].dist = 0
    nodes[0].color = Node.GRAY
    heappush(h, (0, 0))

    while len(h) > 0:
        min_node, min_dist = heappop(h)
        if nodes[min_node].dist < min_dist:
            continue

        nodes[min_node].color = Node.BLACK

        for adj, c in nodes[min_node].adjs:
            if nodes[min_node].dist + c < nodes[adj].dist:
                nodes[adj].dist = nodes[min_node].dist + c
                nodes[adj].color = Node.GRAY
                heappush(h, (adj, nodes[adj].dist))


def print_dist(nodes):
    for u, n in enumerate(nodes):
        print(u, n.dist)


def main():
    nodes = read_adj()
    shortest_path(nodes)
    print_dist(nodes)


if __name__ == '__main__':
    main()

Solution (Python):
from sys import maxsize
from heapq import heappop, heappush


class Node:
    WHITE = 0
    GRAY = 1
    BLACK = 2

    def __init__(self):
        self.adjs = []
        self.color = Node.WHITE
        self.dist = maxsize

    def __lt__(self, other):
        return self.dist < other.dist

    def __repr__(self):
        return "Node({},{},{})".format(self.adjs, self.color, self.dist)


def read_adj():
    n = int(input())
    nodes = [Node() for _ in range(n)]

    for i in range(n):
        u, _, *rest = [int(x) for x in input().split()]
        for i in range(0, len(rest) - 1, 2):
            nodes[u].adjs.append((rest[i], rest[i+1]))

    return nodes


def shortest_path(nodes):
    h = []

    nodes[0].dist = 0
    nodes[0].color = Node.GRAY
    heappush(h, (0, 0))

    while len(h) > 0:
        min_node, min_dist = heappop(h)
        if nodes[min_node].dist < min_dist:
            continue

        nodes[min_node].color = Node.BLACK

        for adj, c in nodes[min_node].adjs:
            if nodes[min_node].dist + c < nodes[adj].dist:
                nodes[adj].dist = nodes[min_node].dist + c
                nodes[adj].color = Node.GRAY
                heappush(h, (adj, nodes[adj].dist))


def print_dist(nodes):
    for u, n in enumerate(nodes):
        print(u, n.dist)


def main():
    nodes = read_adj()
    shortest_path(nodes)
    print_dist(nodes)


if __name__ == '__main__':
    main()

Solution (Python):
from sys import maxsize
from heapq import heappop, heappush


class Node:
    WHITE = 0
    GRAY = 1
    BLACK = 2

    def __init__(self):
        self.adjs = []
        self.color = Node.WHITE
        self.dist = maxsize

    def __lt__(self, other):
        return self.dist < other.dist

    def __repr__(self):
        return "Node({},{},{})".format(self.adjs, self.color, self.dist)


def read_adj():
    n = int(input())
    nodes = [Node() for _ in range(n)]

    for i in range(n):
        u, _, *rest = [int(x) for x in input().split()]
        for i in range(0, len(rest) - 1, 2):
            nodes[u].adjs.append((rest[i], rest[i+1]))

    return nodes


def shortest_path(nodes):
    h = []

    nodes[0].dist = 0
    nodes[0].color = Node.GRAY
    heappush(h, (0, 0))

    while len(h) > 0:
        min_dist, min_node = heappop(h)

        nodes[min_node].color = Node.BLACK
        if nodes[min_node].dist < min_dist:
            continue

        for adj, c in nodes[min_node].adjs:
            if nodes[adj].color == Node.BLACK:
                continue
            if nodes[min_node].dist + c < nodes[adj].dist:
                nodes[adj].dist = nodes[min_node].dist + c
                nodes[adj].color = Node.GRAY
                heappush(h, (nodes[adj].dist, adj))


def print_dist(nodes):
    for u, n in enumerate(nodes):
        print(u, n.dist)


def main():
    nodes = read_adj()
    shortest_path(nodes)
    print_dist(nodes)


if __name__ == '__main__':
    main()

Solution (Python):


import sys
from heapq import heappush, heappop

sys.setrecursionlimit(10 ** 9)
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
INF=float('inf')


def dijkstra(N: int, nodes: list, src: int) -> list:
    
    
    res = [float('inf')] * N
    
    que = [(0, src)]
    res[src] = 0
    
    while len(que) != 0:
        
        dist, cur = heappop(que)
        
        for nxt, cost in nodes[cur]:
            
            if res[cur] + cost < res[nxt]:
                res[nxt] = res[cur] + cost
                
                heappush(que, (res[nxt], nxt))
    
    return res

N=INT()
nodes=[[] for i in range(N)]
for i in range(N):
    l=LIST()
    u=l[0]
    k=l[1]
    l=l[2:]
    for j in range(0, k*2, 2):
        nodes[u].append((l[j], l[j+1]))

ans=dijkstra(N, nodes, 0)
for i in range(N):
    print(i, ans[i])

Solution (Python):
import sys, itertools
f = sys.stdin

def take2(iterable):
    i = iter(iterable)
    while True:
        yield next(i), next(i)

n = int(f.readline())
adj =[[(v, c) for v, c in take2(map(int, f.readline().split()[2:]))] for _ in range(n)]
d = [100000 * 100 + 1] * n

d[0] = 0
queue ={0}
while len(queue):
    parent = queue.pop()
    for v, c in adj[parent]:
        if d[v] > d[parent] + c:
            d[v] = d[parent] + c
            queue.update([v])

for i in range(n):
    print(i, d[i])

Solution (Python):
import heapq

INFTY = 100000000000000000000000000
heap = []
n = int(raw_input())
graph = {}
for i in range(n):
    entry = map(int, raw_input().split(' '))
    u = entry[0]
    k = entry[1]
    graph[u] = {}
    for j in range(k):
        v = entry[2+j*2]
        c = entry[2+j*2+1]
        graph[u][v] = c

d = {}
visited = {}
heap = []
for i in range(n):
    d[i] = INFTY
    visited[i] = False

s = 0
d[s] = 0
visited[s] = True
heapq.heappush(heap,(0,s))

while len(heap) > 0:
    mincost,minnode = heapq.heappop(heap)
    if mincost > d[minnode]: continue
    visited[minnode] = True
    for i in graph[minnode]:
        if not visited[i] and d[i] > d[minnode] + graph[minnode][i]:
            d[i] = d[minnode] + graph[minnode][i]
            heapq.heappush(heap, (d[i],i))

for i in range(n):
    print i,d[i]

Solution (Python):

from heapq import heappop, heappush
n = int(input())
inf = 100 * 100000

M = [[] for i in range(n)]

for i in range(n):
    L = list(map(int, input().split()))
    u = L[0]
    k = L[1]
    v = L[2::2]
    c = L[3::2]
    for j in range(k):
        M[u].append((v[j], c[j]))
result = [0] * n
S = set()
V = set(range(n))
S.add(0)
d = []
for m in M[0]:
    heappush(d, (m[1], m[0]))
T = V - S

while len(T) != 0:
    c, v = heappop(d)
    while v in S:
        c, v = heappop(d)
    result[v] = c
    S.add(v)
    T.remove(v)
    for m in M[v]:
        if m[0] in T:
            heappush(d, (result[v] + m[1], m[0]))

for i, v in enumerate(result):
    print(i, v)

Solution (Python):
import heapq

n = int(input())
edges = [None] * n
for _ in range(n):
    l = list(map(int, input().split()))
    edges[l[0]] = set((c, j) for j, c in (l[i:i + 2] for i in range(2, l[1] * 2 + 2, 2)))

min_costs = [0] * n
visited = [False] * n
visited[0] = True
queue = list(edges[0])
heapq.heapify(queue)

while queue:
    total_cost, vertex = heapq.heappop(queue)
    if visited[vertex]:
        continue
    visited[vertex] = True
    min_costs[vertex] = total_cost
    for cost, vertex2 in edges[vertex]:
        if not visited[vertex2]:
            heapq.heappush(queue, (total_cost + cost, vertex2))

for r in enumerate(min_costs):
    print(*r)

Solution (Python):
import heapq

n = int(input())
edges = [None] * n
for _ in range(n):
    l = list(map(int, input().split()))
    edges[l[0]] = set((c, j) for j, c in (l[i:i + 2] for i in range(2, l[1] * 2 + 2, 2)))

min_costs = [0] * n
visited = [False] * n
visited[0] = True
queue = list(edges[0])
heapq.heapify(queue)
remains = n - 1

while True:
    total_cost, vertex = heapq.heappop(queue)
    if visited[vertex]:
        continue
    visited[vertex] = True
    min_costs[vertex] = total_cost
    remains -= 1
    if not remains:
        break
    for cost, vertex2 in edges[vertex]:
        if not visited[vertex2]:
            heapq.heappush(queue, (total_cost + cost, vertex2))

for r in enumerate(min_costs):
    print(*r)

Solution (Python):
import heapq
from collections import deque

WHITE = 0
GRAY = 1
BLACK = 2

n = int(raw_input())
M = []
for i in xrange(n):
    M.append(deque([]))
color = [WHITE] * n
d = [float('inf')] * n

for i in xrange(n):
    a = map(int, raw_input().split())
    for j in xrange(a[1]):
        M[a[0]].append((a[2*(j+1)],a[2*(j+1)+1]))

def dijkstra():
    d[0] = 0

    PQ = []
    heapq.heappush(PQ,(0,0))
    
    color[0] = GRAY

    while len(PQ) > 0:
        f = heapq.heappop(PQ)
        u = f[1]
        color[u] = BLACK

        if d[u] < f[0]:
            continue

        for j in xrange(len(M[u])):
            v = M[u][j][0]
            if color[v] == BLACK:
                continue
            if d[v] > d[u] + M[u][j][1]:
                d[v] = d[u] + M[u][j][1]
                heapq.heappush(PQ, (d[v],v))
                color[v] = GRAY

    for i in xrange(n):
        print "{} {}".format(i, d[i])

    return 0

def main():
    dijkstra()
    return 0

main()