Go: package main

import (
	"bufio"
	"container/heap"
	"fmt"
	"os"
	"strconv"
)

//make priority queue
type Item struct {
	id       int
	priority int // this state is priority and distans from start
	index    int
}

type PriorityQueue []*Item

func (pq PriorityQueue) Len() int { return len(pq) }

// we want the ascending-priority-queue.
// So we use less than here.
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 = j
	pq[j].index = i
}

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
}

func (pq *PriorityQueue) update(dist int, id int) {
	for _, i := range *pq {
		if i.id == id {
			if i.priority > dist {
				i.priority = dist
			}
		}
	}
}

// the main code
func nextInt(sc *bufio.Scanner) int {
	sc.Scan()
	n, _ := strconv.Atoi(sc.Text())
	return n
}

func dikstra(dist *[][]int, n int) *[]int {
	// preparate priority queue
	inf := 1 << 30
	pq := make(PriorityQueue, n)
	shortests := make([]int, n)
	for i := 0; i < n; i++ {
		pq[i] = &Item{
			id:       i,
			priority: inf,
			index:    i,
		}

		shortests[i] = inf
	}
	pq[0].priority = 0
	shortests[0] = 0
	heap.Init(&pq)

	for pq.Len() > 0 {
		item := heap.Pop(&pq).(*Item)
		shortests[item.id] = item.priority
		for id, d := range (*dist)[item.id] {
			if d != -1 {
				pq.update(item.priority+d, id)
			}
		}
		heap.Init(&pq)
	}
	return &shortests
}

func main() {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)

	// init
	n := nextInt(sc)
	dist := make([][]int, n)
	for i := 0; i < n; i++ {
		dist[i] = make([]int, n)
		for j := 0; j < n; j++ {
			dist[i][j] = -1
		}
	}

	// std Input
	for i := 0; i < n; i++ {
		u := nextInt(sc)
		k := nextInt(sc)
		for j := 0; j < k; j++ {
			v, c := nextInt(sc), nextInt(sc)
			dist[u][v] = c
		}
	}

	shortests := dikstra(&dist, n)
	for i, d := range *shortests {
		fmt.Printf("%d %d\n", i, d)
	}
}

C++: // #include {{{
#include <array>
#include <vector>
#include <deque>
#include <queue>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <string>
#include <algorithm>
#include <numeric>
#include <functional>
#include <iterator>
#include <tuple>
#include <utility>
#include <random>
#include <limits>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cstdint>
#include <cassert>
// }}}

using namespace std;

// macros {{{
template<typename T, size_t N>
constexpr size_t NELEMS(T (&)[N]) { return N; }

template<typename InputIt>
void PRINT_RANGE(InputIt first, InputIt last)
{
    for(; first != last; ++first)
        cout << *first << (first==last ? "" : " ");
    cout << "\n";
}

template<typename T>
void PRINT_MATRIX(T mat, size_t nr, size_t nc)
{
    for(size_t r = 0; r < nr; ++r) {
        for(size_t c = 0; c < nc; ++c) {
            cout << mat[r][c] << (c==nc-1 ? "" : " ");
        }
        cout << "\n";
    }
}

#define FOR(i, start, end) for(int i = (start); i < (end); ++i)
#define REP(i, n) FOR(i, 0, n)
// }}}

// types {{{
using ll  = long long;
using ull = unsigned long long;
// }}}

struct Edge {
    int to;
    int cost;
    Edge(int to_arg, int cost_arg) : to(to_arg), cost(cost_arg) {}
};

constexpr int INF = numeric_limits<int>::max();

int N;
vector<Edge> G[10000];

void solve()
{
    vector<int> cost(N);
    deque<bool> done(N);

    fill(begin(cost), end(cost), INF);
    cost[0] = 0;

    for(;;) {
        int mincost = INF;
        int minv    = -1;

        REP(v, N) {
            if(done[v]) continue;
            if(cost[v] < mincost) {
                mincost = cost[v];
                minv    = v;
            }
        }

        if(minv == -1) break;

        done[minv] = true;

        for(const Edge& e : G[minv]) {
            if(done[e.to]) continue;
            int cost_new = cost[minv] + e.cost;
            if(cost_new < cost[e.to]) {
                cost[e.to] = cost_new;
            }
        }
    }

    REP(v, N) {
        cout << v << " " << cost[v] << "\n";
    }
}

int main()
{
    cin >> N;
    REP(_, N) {
        int from, k;
        cin >> from >> k;
        REP(_, k) {
            int to, cost;
            cin >> to >> cost;
            G[from].emplace_back(to, cost);
        }
    }

    solve();

    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 <bits/stdc++.h>
using ll = long long;
using namespace std;
static const int MAX = 1e4;
static const int INFTY = (1<<20);
static const int WHITE = 0;
static const int GRAY = 1;
static const int BLACK = 2;

int n;
vector<pair<int,int> > adj[MAX];

void dijkstra(){
  priority_queue<pair<int,int> > PQ;
  int color[MAX];
  int d[MAX];
  for(int i=0;i<n;i++){
    d[i] = INFTY;
    color[i] = WHITE;
  }
  d[0] = 0;
  PQ.push(make_pair(0,0));
  color[0] = GRAY;

  while(!PQ.empty()){
    pair<int,int> f = PQ.top(); PQ.pop();
    int u = f.second;
    color[u] = BLACK;

    //最小値を取り出し、それが最短でなければ無視
    if(d[u] < f.first * (-1)) continue;

    for(int j=0;j<adj[u].size();j++){
      int v = adj[u][j].first;
      if(color[v] == BLACK) continue;
      if(d[v] > d[u] + adj[u][j].second){
        d[v] = d[u] + adj[u][j].second;
        PQ.push(make_pair(d[v] * (-1), v));
        color[v] = GRAY;
      }
    }
  }
  for(int i=0;i<n;i++){
    cout << i << " " << (d[i]==INFTY ? -1 : d[i]) << endl;
  }
}

int main(){
  int k,u,v,c;
  cin >> n;
  for(int i=0;i<n;i++){
    cin >> u >> k;
    for(int j=0;j<k;j++){
      cin >> v >> c;
      adj[u].push_back(make_pair(v,c));
    }
  }
  dijkstra();
}