Go: package main

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

const (
	white int = iota
	gray
	black
)

type node struct {
	id     int
	parent int
	edge   []int
	weight []int
	color  int
}

type graph struct {
	nodes    []node
	distance []int
}

var infty = 1 << 32
var null = -1

func constructgraph() *graph {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	n := nextInt(sc)

	g := &graph{make([]node, n), make([]int, n)}

	for i := 0; i < n; i++ {
		id := nextInt(sc)
		edges := nextInt(sc)
		n := node{}
		n.id = id
		n.edge = make([]int, edges)
		n.weight = make([]int, edges)
		n.color = white
		for j := 0; j < edges; j++ {
			to := nextInt(sc)
			w := nextInt(sc)
			n.edge[j] = to
			n.weight[j] = w
		}
		g.nodes[id] = n
	}
	return g
}

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

func (g *graph) preparedijkstra(s int) {
	for i := 0; i < len(g.nodes); i++ {
		g.nodes[i].color = white
		g.nodes[i].parent = null
		g.distance[i] = infty
	}
	g.distance[s] = 0
}

func (g *graph) dijkstra(s int) []int {
	g.preparedijkstra(s)

	for {
		var u int
		mincost := infty
		for i, d := range g.distance {
			if g.nodes[i].color != black && mincost > d {
				mincost = d
				u = i
			}
		}

		if mincost == infty {
			break
		}

		g.nodes[u].color = black

		for i, e := range g.nodes[u].edge {
			if g.nodes[e].color != black {
				if g.distance[e] > g.distance[u]+g.nodes[u].weight[i] {
					g.distance[e] = g.distance[u] + g.nodes[u].weight[i]
					g.nodes[e].parent = u
					g.nodes[e].color = gray
				}
			}
		}
	}
	return g.distance
}

func printShortestPath(d []int) {
	for i, v := range d {
		fmt.Printf("%d %d\n", i, v)
	}
}

func main() {
	g := constructgraph()
	d := g.dijkstra(0)
	printShortestPath(d)
}

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;

}

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

#define MAX_SIZE 110
#define INF (1<<30)
#define NOT_VISITED 0
#define VISITED 1
#define COMPLETE 2

int n, G[MAX_SIZE][MAX_SIZE];
int d[MAX_SIZE], status[MAX_SIZE];

void dijkstra() {
    for (int i = 0; i < n; i++) {
        d[i] = INF;
        status[i] = NOT_VISITED;
    }
    d[0] = 0;

    int u, min_cost;
    while (1) {
        u = -1;
        min_cost = INF;
        for (int i = 0; i < n; i++) {
            if (min_cost > d[i] && status[i] != COMPLETE) {
                u = i;
                min_cost = d[i];
            }
        }
        if (u == -1) break;
        status[u] = COMPLETE;
        for (int i = 0; i < n; i++) {
            if (status[i] == COMPLETE || G[u][i] == INF) continue;
            if (d[i] > d[u] + G[u][i]) {
                d[i] = d[u] + G[u][i];
                status[i] = VISITED;
            }
        }
    }
    
    return;
}

void print() {
    for (int i = 0; i < n; i++) {
        cout << i << " ";
        if (d[i] == INF) cout << -1;
        else cout << d[i];
        cout << endl;
    }
    return;
}

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

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

    dijkstra();

    print();

    return 0;
}