Go: package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

const BIG_COST = 200000

type Path struct {
	node int
	cost int
}

func main() {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)

	nNodes := scanInt(sc)

	// 各頂点につながる頂点一覧を得るための隣接リスト
	graph := make([][]Path, nNodes)

	for i := 0; i < nNodes; i++ {
		u := scanInt(sc)
		k := scanInt(sc)
		graph[i] = make([]Path, k)
		for d := 0; d < k; d++ {
			graph[u][d] = Path{
				node: scanInt(sc),
				cost: scanInt(sc),
			}
		}
	}

	// 0 から各頂点への最小コストを保存する配列
	minCosts := make([]int, nNodes)

	minCosts[0] = 0
	for i := 1; i < nNodes; i++ {
		minCosts[i] = BIG_COST
	}

	// 各頂点が調査済みかどうかを保持する配列
	checked := make([]bool, nNodes)

	// 最小コストのパスを高速に取得するためのヒープ。
	// コストが小さいものほどルートに来るようにする。
	heap := NewHeap(func(a, b Path) bool {
		return a.cost < b.cost
	})

	heap.insert(Path{node: 0, cost: 0})
	for heap.size > 0 {
		u := heap.extractMax()

		// 隣接行列を使っていた時とは違い、ヒープには
		// 同じノードへのパスが複数含まれうる。そのため既によりコストの低い
		// パスが見つかっている場合は無視する。
		if minCosts[u.node] < u.cost {
			continue
		}

		checked[u.node] = true

		for _, v := range graph[u.node] {
			cost := v.cost + u.cost
			if !checked[v.node] && cost < minCosts[v.node] {
				minCosts[v.node] = cost
				heap.insert(Path{node: v.node, cost: cost})
			}
		}
	}

	w := bufio.NewWriter(os.Stdout)
	for i, cost := range minCosts {
		fmt.Fprintf(w, "%d %d\n", i, cost)
	}
	w.Flush()
}

func scanInt(sc *bufio.Scanner) int {
	sc.Scan()
	v, _ := strconv.Atoi(sc.Text())
	return v
}

// --- From ALDS1_9_C: Priority Queue

type Heap struct {
	size   int
	heap   []Path
	isHigh func(a, b Path) bool
}

func NewHeap(isHigh func(a, b Path) bool) Heap {
	heap := make([]Path, 1) // 先頭は不使用
	return Heap{size: 0, heap: heap, isHigh: isHigh}
}

func (h *Heap) insert(v Path) {
	h.size += 1
	if h.size > len(h.heap)-1 {
		h.heap = append(h.heap, v)
	} else {
		h.heap[h.size] = v
	}

	for i := h.size; i > 1; {
		parent := i / 2
		if h.isHigh(h.heap[parent], h.heap[i]) {
			break
		}
		h.heap[i], h.heap[parent] = h.heap[parent], h.heap[i]
		i = parent
	}
}

func (h *Heap) extractMax() Path {
	if h.size <= 0 {
		panic("cannot extract from empty heap")
	}

	max := h.heap[1]
	h.heap[1] = h.heap[h.size]
	h.size -= 1
	h.maxHeapify(1)

	return max
}

func (h *Heap) maxHeapify(idx int) {
	left := idx * 2
	right := left + 1
	largest := idx

	if left <= h.size {
		if h.isHigh(h.heap[left], h.heap[idx]) {
			largest = left
		}
		if right <= h.size && h.isHigh(h.heap[right], h.heap[largest]) {
			largest = right
		}
	}

	if largest != idx {
		h.heap[idx], h.heap[largest] = h.heap[largest], h.heap[idx]
		h.maxHeapify(largest)
	}
}

Python: import heapq as hp

INF = 1000000

def diikstraMethd(G, s):
    # ?????????
    n = len(G)
    d = [INF for _ in range(n)]
    d[s] = 0
    PQ = [(d[s], s)]

    while PQ:
        u = hp.heappop(PQ)[1]
        for v, c in G[u]:
            alt = d[u] + c
            if d[v] > alt:
                d[v] = alt
                hp.heappush(PQ, (alt, v))

    return d


if __name__ == '__main__':
    n = int(input())
    G = [ [] for _ in range(n)]
    for _ in range(n):
        tmp = list(map(int, input().split()))
        u = tmp.pop(0)
        k = tmp.pop(0)
        for i in range(k):
            v = tmp.pop(0)
            c = tmp.pop(0)
            G[u].append((v, c))

    d = diikstraMethd(G, 0)

    for i in range(n):
        print('{0} {1}'.format(i, d[i]))

Python: from heapq import heapify, heappush, heappop

# input
n = int(input())

nodes = [[] * n for i in range(n)]
for _ in range(n):
    line = list(map(int, input().split()))
    u = line.pop(0)
    k = line.pop(0)
    for i in range(k):
        v = line.pop(0)
        c = line.pop(0)
        nodes[u].append((v, c))


# 変数定義
inf = 100 * 100000 + 1

d = [inf] * n
V = set([i for i in range(n)]) #nodeの一覧
S = set() #探索済みnodeの一覧

# 探索
start = 0
target = start
d[start] = 0

H = []

#for i in range(n):
#    heappush(H, (i, d[i]))
heappush(H, (start, d[start]))

while V != S:
    u, c = heappop(H)
    S.add(u)
    for v, c in nodes[u]:
        if d[u] + c < d[v]:
            d[v] = d[u] + c
            heappush(H, (v, d[v]))
    
# 出力
for i in range(n):
    print(f"{i} {d[i]}")

Python: import heapq
class Edge:
    src=-1
    dest=-1
    cost=-1

def Dijkstra(graph,s):
    INF=int(1e9)
    d=[INF]*len(graph)
    d[s]=0
    used=[False]*len(graph)
    que=[]
    heapq.heappush(que,(0,s))
    while len(que)!=0:
        val=heapq.heappop(que)
        v=val[1]
        if used[v] : continue
        used[v]=True
        for i in range(0,len(graph[v])):
            e=graph[v][i]
            if d[v]+e.cost<d[e.dest]:
                d[e.dest]=d[v]+e.cost
                heapq.heappush(que,(d[e.dest],e.dest))
    return d

V=int(input())
graph=[[] for i in range(0,V)]
for i in range(0,V):
    vals=list(map(int,input().split(" ")))
    k=vals[1]
    for j in range(0,k):
        v=vals[2+j*2]
        c=vals[3+j*2]
        e=Edge()
        e.src=i
        e.dest=v
        e.cost=c
        graph[i].append(e)

d=Dijkstra(graph,0)
for i in range(0,V):
    print(str(i)+" "+str(d[i]))

Python: import sys
readline = sys.stdin.readline
from heapq import heapify, heappush, heappop
INF = float("inf")
def MAIN():
    n = int(input())
    G = [[i, INF] for i in range(n)]
    G[0][1] = 0
    m = {}
    for _ in range(n):
        A = list(map(int, readline().split()))
        m[A[0]] = {}
        for i in range(2, len(A), 2):
            m[A[0]][A[i]] = A[i + 1]
    dp = [(0, 0)]
    while dp:
        cost, u = heappop(dp)
        for v, c in m[u].items():
            if G[v][1] > G[u][1] + c:
                G[v][1] = G[u][1] + c
                heappush(dp, (G[v][1], v))
    print(*G)
MAIN()

Python: from heapq import*
I=float("inf")
def m():
 n=int(input())
 A=[]
 for _ in[0]*n:
  e=list(map(int,input().split()))
  A+=[zip(e[2::2],e[3::2])]
 d=[0]+[I]*n
 H=[(0,0)]
 while H:
  u=heappop(H)[1]
  for v,c in A[u]:
   t=d[u]+c
   if d[v]>t:
    d[v]=t
    heappush(H,(t,v))
 print('\n'.join(f'{i} {d[i]}'for i in range(n)))
m()

Python: import sys, collections, heapq
input = lambda: sys.stdin.readline().rstrip() 
sys.setrecursionlimit(10**7)
INF = 10**10
def I(): return int(input())
def F(): return float(input())
def SS(): return input()
def LI(): return [int(x) for x in input().split()]
def LI_(): return [int(x)-1 for x in input().split()]
def LF(): return [float(x) for x in input().split()]
def LSS(): return input().split()

def dijkstra(G, s):
    que = []
    d = [INF]*len(G)
    d[s] = 0
    heapq.heappush(que, (d[s], s))

    while que:
        p = heapq.heappop(que)
        v = p[1]
        if d[v] < p[0]: continue
        for e in G[v]:
            if d[e[0]] > d[v] + e[1]:
                d[e[0]] = d[v] + e[1]
                heapq.heappush(que, (d[e[0]], e[0]))
    return d

def resolve():
    n = I()
    G = [[] for _ in range(n)]
    for i in range(n):
        ukvc = LI()
        u = ukvc[0]
        k = ukvc[1]
        for j in range(k):
            v = ukvc[2*j+2]
            c = ukvc[2*j+3]
            G[u].append((v, c))

    ans = dijkstra(G, 0)
    for i, e in enumerate(ans):
        print(i, e)

if __name__ == '__main__':
    resolve()