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 sys
import heapq
from collections import namedtuple
inf = float('inf')
Node = namedtuple('Node', 'd, no')

def debug(x, table):
    for name, val in table.items():
        if x is val:
            print('DEBUG:{} -> {}'.format(name, val), file=sys.stderr)
            return None

def SSSP(n, Adj):
    d = [inf] * n
    p = [-1] * n
    checked = [False] * n
    d[0] = 0
    checked[0] = True
    heap = [Node(d[0], 0)]

    while heap:
        u = heapq.heappop(heap)

        if d[u.no] < u.d:
            continue

        checked[u.no] = True

        for v, cost in Adj[u.no]:
            if not checked[v] and d[u.no] + cost < d[v]:
                d[v] = d[u.no] + cost
                p[v] = u.no
                heapq.heappush(heap, Node(d[v], v))

    return d

def solve():
    n = int(input())
    Adj = [[] for i in range(n)]

    for i in range(n):
        u, k, *line = [int(i) for i in input().split()]
        for j in range(k):
            Adj[u].append((line[2*j], line[2*j + 1]))

    # debug(Adj, locals())

    d = SSSP(n, Adj)

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

if __name__ == '__main__':
    solve()

Python: import sys

f_i = sys.stdin

n = int(f_i.readline())

adj = [[100001 for j in range(n)] for i in range(n)]

for x in f_i:
    x = list(map(int, x.split()))
    u = x[0]
    for v, c in zip(x[2::2], x[3::2]):
        adj[u][v] = c

unvisited = [v for v in range(n)]

distance = [999900001] * n
distance[0] = 0

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

import heapq

vcost = [VCost(0, 0)]

def dijkstra():
    while vcost:
        uc = heapq.heappop(vcost)

        if uc.cost > distance[uc.v]:
            continue
        
        unvisited.remove(uc.v)
        
        for v in unvisited:
            weight_uv = adj[uc.v][v]
            if weight_uv != 100001:
                t_cost = uc.cost + weight_uv
                if t_cost < distance[v]:
                    distance[v] = t_cost
                    heapq.heappush(vcost, VCost(v, t_cost))



dijkstra()

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

Python: V=int(input())

G=[[] for i in range(V)]
d=[1001001001 for i in range(V)]

import heapq

def dijkstra(s):
    q=[]
    d[s]=0
    heapq.heappush(q,(0, s))
    
    while(len(q)):
        
        p=heapq.heappop(q)
        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(q, (d[e[0]], e[0]))

import sys
for l in sys.stdin:
    l=list(map(int,l.split()))
    u=l[0]
    k=l[1]
    G[u]=zip(*[iter(l[2:])]*2)
    
dijkstra(0)
for i,c in enumerate(d):
    print(i, c)

Python: #!/usr/bin/env python
from __future__ import division, print_function
from sys import stdin, maxint
from heapq import heappush, heappop, heapify


def main():
    num = int(stdin.readline())

    edges = []
    for _ in xrange(num):
        it = iter(int(s) for s in stdin.readline().split()[1:])
        edges.append([(next(it), next(it)) for _ in xrange(next(it))])

    distances = [[maxint, v] for v in xrange(num)]
    distances[0][0] = 0
    V = set(xrange(1, num))

    if num == 1000:
        solve1000(edges, distances, V)
    else:
        solve(edges, distances, V)

    for d, i in distances:
        print(i, d)


def solve1000(edges, distances, V):
    candidate = []
    mark = set()
    index = 0

    while V:
        for end, cost in edges[index]:
            if end not in V:
                continue
            cost += distances[index][0]
            if distances[end][0] > cost:
                distances[end][0] = cost
                if end not in mark:
                    heappush(candidate, distances[end])
                    mark.add(end)
                else:
                    heapify(candidate)

        index = heappop(candidate)[1]
        mark.remove(index)
        V.discard(index)


def solve(edges, distances, V):
    candidate = []
    index = 0

    while V:
        for end, cost in edges[index]:
            if end not in V:
                continue
            cost += distances[index][0]
            if distances[end][0] > cost:
                distances[end][0] = cost
                heappush(candidate, distances[end])

        index = heappop(candidate)[1]
        V.discard(index)


main()