Go: package main

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

var sc = bufio.NewScanner(os.Stdin)

const T = 10000000

var n int
var A [101][101]int

func nextInt() int {
	sc.Scan()
	i, _ := strconv.Atoi(sc.Text())
	return i
}
func Shortestsearch() {
	var Minimum int
	var d, color [100]int
	var u int

	for i := 0; i < n; i++ {
		d[i] = T
		color[i] = 0
	}

	d[0] = 0
	color[0] = 1

	for {
		Minimum = T
		u = -1
		for i := 0; i < n; i++ {
			if Minimum > d[i] && color[i] != 2 {
				u = i
				Minimum = d[i]
			}
		}
		if u == -1 {
			break
		}
		color[u] = 2
		for j := 0; j < n; j++ {
			if color[j] != 2 && A[u][j] != T {
				if d[j] > d[u]+A[u][j] {
					d[j] = d[u] + A[u][j]
					color[j] = 1
				}
			}
		}
	}
	for i := 0; i < n; i++ {
		fmt.Printf("%d %d\n", i, d[i])
	}
}
func main() {
	sc.Split(bufio.ScanWords)
	n = nextInt()

	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			A[i][j] = T
		}
	}
	for i := 0; i < n; i++ {
		u := nextInt()
		k := nextInt()
		for j := 0; j < k; j++ {
			v := nextInt()
			c := nextInt()
			A[u][v] = c
		}
	}
	Shortestsearch()
}

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

using namespace std; 

int main(void) {
	int u,noe,nov;
	cin >> nov;
	vector< vector< pair<int,int> > > el(nov);
	for (int i = 0; i < nov; ++i) {
		cin >> u >> noe;
		el[u] = vector< pair<int,int> >(noe);
		for (int j = 0; j < noe; ++j) {
			cin >> el[u][j].first;
			cin >> el[u][j].second;
		}
	}

	vector<int> res(nov,(1 << 28));
	res[0] = 0;
	pair<int,int> now,next;
	priority_queue< pair<int,int> > q;
	q.push({0,0});
	while (!q.empty()) {
		now = q.top();
		q.pop();
		for (int i = 0; i < el[now.second].size(); ++i) {
			next = el[now.second][i];
			if (res[next.first] > now.first+next.second) {
				res[next.first] = now.first+next.second;
				q.push({res[next.first],next.first});
			}
		}
	}

	for (int i = 0; i < nov; ++i) {
		cout << i << " ";
		cout << res[i] << endl;
	}

	return 0;
}

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

using namespace std;

int main() {

	int INF = 10000000;

	int n;
	scanf("%d", &n);

	vector<pair<int, int>> G[n];
	for (int i = 0; i < n; ++i) {
		int from;
		scanf("%d", &from);

		int v;
		scanf("%d", &v);

		for (int j = 0; j < v; ++j) {
			int to;
			int w;
			scanf("%d %d", &to, &w);
			G[from].push_back(make_pair(to, w));
		}
	}

	int C[n];
	for (int i = 0; i < n; ++i)
		C[i] = INF;

	int color[n];
	int WHITE = 0;
	int BLACK = 1;
	for (int i = 0; i < n; ++i)
		color[i] = WHITE;

	priority_queue<pair<int, int>> PQ;
	PQ.push(make_pair(0, 0));
	C[0] = 0;
	while (!PQ.empty()) {
		int u = PQ.top().second;
		PQ.pop();
		color[u] = BLACK;
		for (int i = 0; i < (int) G[u].size(); ++i) {
			auto c = G[u][i];
			if (color[c.first] != BLACK && c.second + C[u] < C[c.first]) {
				C[c.first] = c.second + C[u];
				PQ.push(make_pair(C[c.first] * -1, c.first));
			}
		}
	}

	for (int i = 0; i < n; ++i) {
		cout << i << " " << C[i] << endl;
	}

	return 0;
}

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

int main() {
    int n;
    cin >> n;

    int a[n][n];
    int done[n]; // -1: まだ ソレ以外: 確定距離
    for (int i = 0; i < n; ++i) {
        done[i] = -1;
        for (int j = 0; j < n; ++j) {
            a[i][j] = -1;
        }
    }
    for (int i = 0; i < n; ++i) {
        int u, k;
        cin >> u >> k;
        for (int j = 0; j < k; ++j) {
            int v, c;
            cin >> v >> c;
            a[i][v] = c;
        }
    }

    // デバッグ用
//     for (int i = 0; i < n; ++i) {
//         for (int j = 0; j < n; ++j) {
//             cout << i << " " << j << " " << a[i][j] << endl;
//         }
//     }

    // 最初は0が確定済みのノードとなる
    done[0] = 0;

    // 確定済みのノードから行ける、未到達の最小のノードを探す
    // その最小のノードは次の確定済みとする...
    // とりあえずデバッグがてら10回す (最終的にwhileにして条件で対応する)
    // 全てdoneになったら終了
    for (int r = 0; r < n-1; ++r) {
        int next_node = -1;
        int next_cost = 1000000;
        for (int i = 0; i < n; ++i) {
            if (done[i] == -1) {
                // 未確定であれば何もしない
                continue;
            }

            for (int j = 0; j < n; ++j) {
                // 到達済みは見ない
                if (done[j] != -1) {
                    continue;
                }

                // コストが-1なら到達できない
                if (a[i][j] == -1) {
                    continue;
                }

                // 行けるところのコスト (iまでの距離+そこからの距離) を探す
//                cout << "src:" << i << " dst:" << j << " cost:" << a[i][j] << " total:" << done[i] + a[i][j] << endl;

                // 今のnext_costよりもtotalcostが小さければ、次のノードにしたい
                if ((done[i] + a[i][j]) < next_cost) {
                    next_cost = done[i] + a[i][j];
                    next_node = j;
                }
            }
        }
//        cout << "iter:" << r << endl;
//        cout << "next_node:" << next_node << endl;
//        cout << "next_cost:" << next_cost << endl;
        done[next_node] = next_cost;
    }

    for (int i = 0; i < n; ++i) {
        cout << i << " " << done[i] << endl;
    }

    return 0;
}

C++: #include<stdio.h>
#define MAX 100
#define INFTY (2<<21)
#define WHITE 0
#define GRAY 1
#define BLACK 2

int M[MAX][MAX];
int d[MAX];
int color[MAX];

void dijkstra(int n) {
    int u, mincost;
    d[0] = 0;
    color[0] = GRAY;

    while (true) {
        mincost = INFTY;
        u = -1;

        for (int i=0; i<n; i++) {
            if (color[i]!=BLACK && d[i]<mincost) {
                u = i;
                mincost = d[i];
            }
        }

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

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

int main() {
    int n, k, v, c, u;
    scanf("%d", &n);

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

    for (int i=0; i<n; i++) {
        scanf("%d %d", &u, &k);
        for (int j=0; j<k; j++) {
            scanf("%d %d", &v, &c);
            M[u][v] = c;
        }
    }

    dijkstra(n);

    for (int i=0; i<n; i++) {
        printf("%d %d\n", i, d[i]);
    }

    return 0;
}

C++: #include<iostream>
#include<vector>
using namespace std;
int map[100][100];
int n;
int d[100];
int color[100];
int INF = 1000000;
void solve() {
	int all = 1;
	d[0] = 0;
	color[0] = 1;
	vector<int>v;
	v.push_back(0);
	while (all != n)
	{
		int min = INF;
		int k;
		for (int i = 0; i < v.size(); i++) {
			for (int j = 0; j < n; j++) {
				if (!color[j] && d[v[i]] + map[v[i]][j] <= d[j]) {
					d[j] = d[v[i]] + map[v[i]][j];
					if (d[j] < min) {
						min = d[j];
						k = j;
					}
				}
			}
		}
		color[k] = 1;
		v.push_back(k);
		all++;
	}
}
int main(void) {
	cin >> n;
	for (int i = 0; i < n; i++) {
		d[i] = INF;
		for (int j = 0; j < n; j++) map[i][j] = INF;
	}
	for (int i = 0; i < n; i++) {
		int k,all;
		cin >> k>>all;
		for (int j = 0; j < all; j++) {
			int a, b;
			cin >> a >> b;
			map[k][a] = b;
		}
	}
	solve();
	for (int i = 0; i < n; i++) {
		cout << i<<" "<<d[i]<<endl;
	}
}