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)
	}
}

C++: #include <bits/stdc++.h>

using namespace std;

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

#define FOR(i, n, m) for(ll (i)=(m);(i)<(n);++(i))
#define REP(i, n) FOR(i,n,0)
#define OF64 std::setprecision(10)

const ll MOD = 1000000007;
const ll INF = (ll) 1e15;

struct Vertex {
    vector<pll> node;
};

Vertex V[105];
ll C[105];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N;
    cin >> N;
    REP(i, N) {
        ll n, k;
        cin >> n >> k;
        REP(_, k) {
            ll v, c;
            cin >> v >> c;
            V[n].node.push_back(pll(v, c));
        }
    }
    REP(i, N) {
        C[i] = INF;
    }
    priority_queue<pll, vector<pll>, function<bool(pll, pll)>> q([](pll a, pll b) { return a.second > b.second; });
    q.push(pll(0, 0));
    C[0] = 0;
    while (!q.empty()) {
        pll t = q.top();
        q.pop();
        if (C[t.first] < t.second)
            continue;
        for (auto &n : V[t.first].node) {
            ll nxt = n.first;
            ll cost = n.second + t.second;
            if (C[nxt] <= cost)
                continue;
            C[nxt] = cost;
            q.push(pll(nxt, cost));
        }
    }

    REP(i, N) {
        cout << i << " " << C[i] << endl;
    }
    return 0;
}

C++: #include<iostream>
using namespace std;

static const int MAX = 100; //頂点数の最大値100まで対応可能
static const int INFTY = (1<<21);   //適当な値で\inftyを表す
static const int WHITE = 0; //未訪問の頂点を白色で表す
static const int GRAY = 1; //その時点で決定された最短経路木Sに属する
static const int BLACK = 2; //訪問済みの頂点を黒色で表す

int n, M[MAX][MAX]; //Mはuからvへの辺の重みを記録する隣接行列

void dijkstra(){
    int minv;
    int d[MAX], color[MAX];

    for(int i = 0; i < n; i++){
        d[i] = INFTY; //すべての頂点について最短コストを\inftyで初期化しておく（最短経路探索問題だから）
        color[i] = WHITE; //すべての頂点を未訪問としておく
    }

    d[0] = 0; //スタート地点からスタート地点までの距離は0
    color[0] = GRAY; //スタート地点をSに所属させる
    while(1){
        minv = INFTY;
        int u = -1; //最初0の親を-1にしておく
        for(int i = 0; i < n; i++){ 
            if(minv > d[i] && color[i] != BLACK){  //頂点が訪問済みでなく, かつ, スタート地点からiまでの最短コストがminv未満
                u = i; //最短コストを取る点をiへと更新する
                minv = d[i]; //最短コストをd[i]へと更新する
            }
        }
        
        //　V/Sの中でd[u]が最小となる点uの選択が完了した.

        if(u==-1) break;
        color[u] = BLACK; //uを訪問済みにするs
        for(int v = 0; v < n; v++){
            if(color[v] != BLACK && M[u][v] != INFTY){ //vが未訪問の点であり, かつuとvの間にedgeが存在する
                if(d[v] > d[u] + M[u][v]){ //uからedgeを通りvへ向かう距離の方が, それまでのd[v]より小さいならば
                    d[v] = d[u] + M[u][v]; //d[v]の値を更新する
                    color[v] = GRAY; //vを決定された最短経路木Sに属させる
                }
            }
        }
    }

    for(int i = 0; i < n; i++){
        cout << i << " " << (d[i] == INFTY ? -1 : d[i]) << endl; //距離が\inftyか?? 
    }
}

int main(){
    cin >> n;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            M[i][j] = INFTY;
        }
    }

    int k, c, u, v;
    for(int i = 0; i < n; i++){
        cin >> u >> k; //頂点番号を出次数を入力する
        for(int j = 0; j < k; j++){
            cin >> v >> c;
            M[u][v] = c; //有効辺の重みを入力する
        }
    }

    dijkstra(); //Dijkstra法を実施する
    return 0;
}

C++: #include<iostream>
using namespace std;
static const int MAX = 100;
static const int INFTY = (1 << 21);
static const int WHITE = 0;
static const int GRAY = 1;
static const int BLACK = 2;


int n, M[MAX][MAX];

void dijkstra() {

	int minv;
	int d[MAX], color[MAX];

	for (int i = 0; i < n; i++) {
		d[i] = INFTY;
		color[i] = WHITE;
	}

	d[0] = 0;
	color[0] = GRAY;
	while (1)
	{
		minv = INFTY;
		int u = -1;
		for (int i = 0; i < n; i++) {
			if (minv > d[i] && color[i] != BLACK) {
				u = i;
				minv = d[i];
			}
		}

		if (u == -1)break;
		color[u] = BLACK;

		for (int v = 0; v < n; v++) {
			if (color[v] != BLACK && M[u][v] != INFTY) {
				if (d[v] > d[u] + M[u][v]) {
					d[v] = d[u] + M[u][v];
					color[v] = GRAY;
				}
			}
		}

	}

	for (int i = 0; i < n; i++) {
		cout << i << " " << (d[i] == INFTY ? -1 : d[i]) << endl;
	}
}

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			M[i][j] = INFTY;
		}
	}
	int k, c, u, v;
	for (int i = 0; i < n; i++) {
		cin >> u >> k;
		for (int j = 0; j < k; j++) {
			cin >> v >> c;
			M[u][v] = c;
		}
	}

	dijkstra();

	return 0;
}

C++: #include<iostream>

using namespace std;

static const int MAX = 100;
static const int INFTY = (1<<21);
static const int WHITE = 0;
static const int GRAY = 1;
static const int BLACK = 2;

int n, M[MAX][MAX];

void dijkstra(){
  int minv;
  int d[MAX], color[MAX];

  for(int i = 0; i < n; i++){ //??????????????????????????????
    d[i] = INFTY;
    color[i] = WHITE;
  }

  d[0] = 0;
  color[0] = GRAY;
  
  while(1){

    minv = INFTY;
    int u = -1;
    
    for(int i = 0; i < n; i++){
      if(minv > d[i] && color[i] != BLACK){
	u = i;
	minv = d[i];
      }
    }

    if(u == -1)break;

    color[u] = BLACK;

    for(int v = 0; v < n; v++){
      if(color[v] != BLACK && M[u][v] != INFTY){
	if(d[v] > d[u]+M[u][v]){
	  d[v] = d[u]+M[u][v];
	  color[v] = GRAY;
	}
      }      
    }
  }

  for(int i = 0; i < n; i++){
    cout << i << " " << (d[i] == INFTY ? -1 : d[i]) << endl;
  }
}

int main(){
  cin >> n;

  for(int i = 0; i < n; i++){
    for(int j = 0; j < n; j++){
      M[i][j] = INFTY;
    }
  }

  int k, c, u, v;

  for(int i = 0; i < n; i++){
    cin >> u >> k; 
    for(int j = 0; j < k; j++){
      cin >> v >> c;
      M[u][v] = c;
    }
  }

  dijkstra();

  return 0;

}