這個題非常值得一做，儘管我用了很長時間

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<queue>
#define per(i,a,b) for(int i=a;i<b;i++)
#define rep(i,b,a) for(int i=b;i>a;i--)
using namespace std;
const int N=1e4+5;
const int INF=1<<25;
int n,u,v,w;
struct node{
	int v;
	int c;
	node (int _v=0,int _c=0):v(_v),c(_c){}//
	bool operator<(const node &r)const{
		return c>r.c;
	}
};//數越小優先順序越高

struct Edge{
	int v,cost;
	Edge(int _v=0,int _cost=0):v(_v),cost(_cost){};
};//下一個結點和權值

vector<Edge>E[N];
bool vis[N];
int dist[N];
void Dijkstra(int n,int start){
	memset(vis,false,sizeof(vis));
	for(int i=1;i<=n;i++) dist[i]=INF;
		priority_queue<node>que;
		while(!que.empty()) que.pop();//優先佇列置空

		dist[start]=0;
		que.push(node(start,0));

		node tmp;
		while(!que.empty()){
			tmp=que.top();
			que.pop();

			int u=tmp.v;
			if(vis[u]) continue;

			vis[u]=true;
			for(int i=0;i<E[u].size();i++){
				int v=E[u][i].v;
				int cost =E[u][i].cost;
				if(!vis[v]&&dist[u]+cost<dist[v]){//判斷直接和繞過另一個那個進
					dist[v]=dist[u]+cost;
					que.push(node(v,dist[v]));
				}
			}
		}
}
int main(){
	int n,k;
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>u>>k;
		per(i,0,k){
			cin>>v>>w;
			E[u].push_back(Edge(v,w));
		}
	}
	Dijkstra(n,0);
	for(int i=0;i<n;i++)
        cout<<i<<" "<<dist[i]<<endl;
	return 0;
}