Go: package main

import (
	"bufio"
	"container/heap"
	"fmt"
	"os"
)

type Item struct {
	value    int // The value of the item; arbitrary.
	priority int // The priority of the item in the queue.
	// The index is needed by update and is maintained by the heap.Interface methods.
	index int // The index of the item in the heap.
}
type PriorityQueue []*Item

func (pq PriorityQueue) Len() int { return len(pq) }
func (pq PriorityQueue) Less(i, j int) bool {
	return pq[i].priority < pq[j].priority
}
func (pq PriorityQueue) Swap(i, j int) {
	pq[i], pq[j] = pq[j], pq[i]
	pq[i].index = i
	pq[j].index = j
}
func (pq *PriorityQueue) Push(x interface{}) {
	n := len(*pq)
	item := x.(*Item)
	item.index = n
	*pq = append(*pq, item)
}
func (pq *PriorityQueue) Pop() interface{} {
	old := *pq
	n := len(old)
	item := old[n-1]
	item.index = -1 // for safety
	*pq = old[0 : n-1]
	return item
}

type Q struct {
	pq *PriorityQueue
	ds map[int]int // It has always same value as pq. For random access.
}

func newQ() *Q {
	pq := new(PriorityQueue)
	heap.Init(pq)
	return &Q{
		pq: pq,
		ds: make(map[int]int),
	}
}
func (q *Q) pop() (u, d int) {
	item := heap.Pop(q.pq).(*Item)
	delete(q.ds, item.value)
	return item.value, item.priority
}
func (q *Q) get(u int) (int, bool) { // returns d
	d, ok := q.ds[u]
	return d, ok
}
func (q *Q) set(u, d int) {
	q.ds[u] = d
	for _, item := range *q.pq {
		if item.value != u {
			continue
		}

		item.priority = d
		heap.Fix(q.pq, item.index)
		return
	}
	heap.Push(q.pq, &Item{
		priority: d,
		value:    u,
	})
}
func (q *Q) len() int {
	return len(q.ds)
}

func main() {
	r := bufio.NewReader(os.Stdin)

	var n int
	fmt.Fscan(r, &n)
	cs := make([]map[int]int, n) // []v_i -> c_i
	for h := 0; h < n; h++ {
		var u, k int
		fmt.Fscan(r, &u, &k)
		m := make(map[int]int)
		for i := 0; i < k; i++ {
			var v, c int
			fmt.Fscan(r, &v, &c)
			m[v] = c
		}

		cs[u] = m
	}

	q := newQ()
	q.set(0, 0)
	visited := make(map[int]struct{})
	out := make([]int, n)

	for q.len() > 0 {
		u, d := q.pop()
		for uu, c := range cs[u] {
			// Consider the connection between u and uu.
			if _, ok := visited[uu]; ok {
				continue
			}
			dd := d + c
			if d, ok := q.get(uu); !ok || d > dd {
				q.set(uu, dd)
			}
		}

		out[u] = d
		visited[u] = struct{}{}
	}

	for u, d := range out {
		fmt.Println(u, d)
	}
}

Python: # -*- coding: utf-8 -*-
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

    # sample input when given as adjacency list (adj_matrix=False)
    edges = [[(1, 2), (2, 5), (3, 4)],         # node 0
             [(0, 2), (3, 3), (4, 6)],         # node 1
             [(0, 5), (3, 2), (5, 6)],         # node 2
             [(0, 4), (1, 3), (2, 2), (4, 2)], # node 3
             [(1, 6), (3, 2), (5, 4)],         # node 4
             [(2, 6), (4, 4)]]                 # node 5

    # sample input when given as adjacency matrix (adj_matrix=True)
    edges = [[0, 2, 5, 4, inf, inf],           # node 0
             [2, 0, inf, 3, 6, inf],           # node 1
             [5, inf, 0, 2, inf, 6],           # node 2
             [4, 3, 2, 0, 2, inf],             # node 3
             [inf, 6, inf, 2, 0, 4],           # node 4
             [inf, inf, 6, inf, 4, 0]]         # node 5
    '''

    n = len(edges)

    inf = float('inf')
    # costs = [inf] * n
    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 = [inf] * n
    costs = defaultdict(lambda: inf)
    costs[start] = 0
    pq, rem = [(0, start)], n - 1
    # prevs = [-1 for _ in [None] * n]
    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 enumerate(dijkstra(edges, 0)):
        print(i, cost)

Python: import sys

f_i = sys.stdin

n = int(f_i.readline())

class VCost:
    def __init__(self, v, cost):
        self.v_n = v
        self.cost = cost
    def __lt__(self, other):
        return self.cost < other.cost
    def __gt__(self, other):
        return self.cost > other.cost

adj = [[VCost(int(v), int(c)) for v, c in zip(x.split()[2::2], x.split()[3::2])] for x in f_i]

import heapq

def dijkstra():
    PQ = []
    isVisited = [False] * n
    distance = [999900001] * n

    distance[0] = 0
    heapq.heappush(PQ, VCost(0, 0))
    
    while PQ:
        uc = heapq.heappop(PQ)
        u = uc.v_n

        if uc.cost > distance[uc.v_n]:
            continue

        isVisited[u] = True
        
        for vc in adj[u]:
            v = vc.v_n
            if isVisited[v] == True:
                continue
            t_cost = distance[u] + vc.cost
            if t_cost < distance[v]:
                distance[v] = t_cost = distance[u] + vc.cost
                heapq.heappush(PQ, VCost(v, t_cost))

    for v, d in enumerate(distance):
        print(v, d)


dijkstra()