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)
	}
}

C++: #include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<utility>

#define INF 9999999999

using namespace std;

const int MAX_N = 100000;

int d[MAX_N];

struct edge{
	int to;   //?????????
	int cost; //?????????
};

typedef pair<int,int>P;

void dijkstra(int s,int n)
{
	int k,l,p,q,r,u,v,cost,m;
	vector<edge> edgelist[n];
	for(int i=0;i<n;i++){
		scanf("%d",&u);
		scanf("%d",&m);
		for(int j=0;j<m;j++){
			edge e;
			scanf("%d",&v);
			scanf("%d",&cost);
			e.to = v;	e.cost = cost;
			edgelist[i].push_back(e);
		}
	}
	priority_queue<P,vector<P>,greater<P> > que;
	d[s] = 0;
	que.push(P(0,s));
	while(!que.empty()){
		P p = que.top();
		que.pop();
		int v = p.second;
		if(d[v] < p.first) continue;
		vector<edge>::iterator it = edgelist[v].begin();
		while(it != edgelist[v].end()){
			if(d[it->to] > d[v] + it->cost){
				d[it->to] = d[v] + it->cost;
				que.push(P(d[it->to],it->to));
			}
			it++;
		}
	}
}

int main()
{
	int k,l,m,n,s;
	scanf("%d",&n); // ?????°?????\???
	fill(d,d+n,INF);
	dijkstra(0,n);
	for(int i=0;i<n;i++){
		printf("%d %d\n",i,d[i]);
	}
}

C++: #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <climits>
#include <string>
#include <vector>
#include <utility>
#include <functional>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <cassert>

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const ll MOD = ll(1e9) + 7;
const ll INF = LLONG_MAX;
const ll MAX_N = ll(1e5) + 5;
const double EPS = 10e-8;
const double PI = 3.14159265359;

#define REP(i, n) for (ll i = 0; i < n; i++)
#define REPR(i, n) for (ll i = n; i >= 0; i--)
#define FOR(i, m, n) for (ll i = m; i < n; i++)

int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

struct UF {
    static const int MAX_N = 100000;
    int p[MAX_N];
    int r[MAX_N];

    void init(int n) {
        REP(i, n) {
            p[i] = i;
            r[i] = 0;
        }
    }
    int find(int x) {
        if (p[x] == x) return x;
        return p[x] = find(p[x]);
    }
    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) return;
        if (r[x] < r[y]) swap(x, y);
        if (r[x] == r[y]) r[x]++;
        p[y] = x;
    }
    bool same(int x, int y) { return find(x) == find(y); }
};

int n, d[100000];
vector<P> g[100000];

void dijkstra(int s) {
    fill(d, d + n, 99999999*2);
    d[s] = 0;
    priority_queue<P> que;
    que.push({0, s});
    while (!que.empty()) {
        P p = que.top();
        que.pop();
        int u = p.second;
        if (p.first > d[u]) continue;
        REP(i, g[u].size()) {
            int v = g[u][i].first, cost = g[u][i].second;
            if (d[v] > d[u] + cost) {
                d[v] = d[u] + cost;
                que.push({d[v], v});
            }
        }
    }
}

int main() {
    cin >> n;
    REP(i, n) {
        int from, k;
        cin >> from >> k;
        REP(j, k) {
            int to, c;
            cin >> to >> c;
            g[from].push_back({to, c});
        }
    }
    dijkstra(0);
    REP(i, n) { cout << i << " " << d[i] << endl; }
    return 0;
}