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++: //Dijkstra
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
typedef pair<int, int> P;
int n, dis[10001], n_left;
vector< vector<P> > G(10001);
priority_queue< P, vector<P>, greater<P> > heap;
bool visited[10001];

int main(){
    int u, v, k, w;

    scanf("%d", &n);
    n_left = n;
    for(int i=0; i<n; i++){
        scanf("%d %d", &u, &k);
        for(int j=0; j<k; j++){
            scanf("%d %d", &v, &w);
            G[u].push_back(P(w, v));
        }
    }
    memset(visited, 0, sizeof(visited));
    memset(dis, 127, sizeof(dis));

    dis[0] = 0;
    heap.push(P(0, 0));
    while(!heap.empty() && n_left>0){
        P top = heap.top();
        heap.pop();
        if(!visited[top.second]){
            u = top.second;
            visited[u] = true;
            --n_left;
            for(int i=0; i<G[u].size(); i++){
                v = G[u][i].second;
                w = dis[u] + G[u][i].first;
                if(!visited[v] && w < dis[v]){
                    heap.push(P(w, v));
                    dis[v] = w;
                }
            }
        }
    } 
    for(int i=0; i<n; i++){
        printf("%d %d\n", i, dis[i]);
    }
    return 0;
}

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

#define INF 2000000000

typedef pair<int, int> P;
struct edge{ int to, cost;};

int n;
vector<edge> G[10000];
int d[10000];

void dijkstra(int s){
  
  priority_queue<P, vector<P>, greater<P> > que;
  
  for(int i = 0;i < n;i++) d[i] = INF;
  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;
    for(int i = 0;i < G[v].size();i++){
      edge e = G[v][i];
      if(d[e.to] > d[v] + e.cost){
        d[e.to] = d[v] + e.cost;
        que.push(P(d[e.to], e.to));
      }
    }
  }
}

int main(){
  
  int u, k;
  edge e;
  
  cin >> n;
  for(int i = 0;i < n;i++){
    cin >> u >> k;
    for(int j = 0;j < k;j++){
      cin >> e.to >> e.cost;
      G[u].push_back(e);
    }
  }
  
  dijkstra(0);
  
  for(int i = 0;i < n;i++) cout << i << " " << d[i] << endl;
  
  return 0;
}

C++: // ALDS1_12_C.cpp
// Graph II - Single Source Shortest Path II
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <queue>
using namespace std;

struct Node {
    int v;
    int c;
    Node(int v, int c): v(v), c(c) {}
};

const int maxN = 10000;
static const int WHITE = 0;
static const int GRAY = 1;
static const int BLACK = 2;
static const int INF = (1<<21);
vector<Node> G[maxN];
vector<int> dist(maxN, INF);
vector<int> p(maxN);
vector<int> color(maxN, WHITE);
priority_queue<pair<int, int>> mq;
int n;

void dijkstra(int x)
{
    dist[x] = 0;
    p[x] = -1;
    mq.push({-dist[x], x});
    while(!mq.empty()) {
        pair<int, int> P = mq.top();
        mq.pop();
        int mindist = -P.first;
        int u = P.second;
        if (color[u] == BLACK) continue;

        for (int i = 0; i < G[u].size(); i++) {
            if (color[u] == BLACK) continue;
            int v = G[u][i].v;
            int c = G[u][i].c;
            if (dist[u] + c < dist[v]) {
                dist[v] = dist[u] + c;
                p[v] = u;
                color[v] = GRAY;
                mq.push({-dist[v], v});
            }
        }
        color[u] = BLACK;
    }
}

int main()
{
    freopen("in7.txt", "r", stdin);
    freopen("out7.txt", "w", stdout);
    cin >> n;
    int u, k;
    int v, c;
    for (int i = 0; i < n; i++) {
        cin >> u >> k;
        for (int j = 0; j < k; j++) {
            cin >> v >> c;
            G[i].push_back(Node(v, c));
        }
    }
    dijkstra(0);
    for (int i = 0; i < n; i++) {
        printf("%d %d\n", i, dist[i]);
    }
    return 0;
}