PTA A1003 Emergency (Dijkstra single source shortest seek)

Meaning of the questions: Given n cities, m Article undirected edges. Each city has a certain number of rescue teams, while the right side of all known. Now given the start and end, from start to finish seeking the shortest route and the number of the number of rescue teams on the shortest path. If there are multiple shortest paths, the number of output and maximum.

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
const int N=505;
int n,m;//n为顶点数，m为边数 
int G[N][N];//邻接矩阵 
bool vis[N];//标记顶点访问与否 

int d[N];//最短路径 

int num[N];//最短路径条数 

int weight[N];//点权 
int w[N];//最大点权之和 

void Dijkstra(int s){//s为起点 
	
	memset(d,INF,sizeof(d));//起点到其他点的最短路径初始化为INF 
	d[s]=0;//起点到起点自身的最短路径初始化为0 
	w[s]=weight[s];//起点到自身的最大点权之和w[s]初始化为起点的点权weight[s] 
	num[s]=1;//起点到自身的最短路径条数初始化为1 
	
	for(int i=0;i<n;++i){//得把整个图跑完才行吗？是的。 
		int u=-1,MIN=INF;
		for(int j=0;j<n;++j){
			if(!vis[j]&&d[j]<MIN){
				u=j;
				MIN=d[j];
			}
		} 
		if(u==-1) return;
		vis[u]=true;
		for(int v=0;v<n;++v){
			if(!vis[v]&&G[u][v]!=INF){
				if(d[u]+G[u][v]<d[v]){
					d[v]=d[u]+G[u][v];
					w[v]=w[u]+weight[v];
					num[v]=num[u];
				}else if(d[u]+G[u][v]==d[v]){
					num[v]+=num[u];
					if(w[u]+weight[v]>w[v]) w[v]=w[u]+weight[v];
				}				
			}
		}
	}
}
int main(){
	int c1,c2,u,v;
	scanf("%d%d%d%d",&n,&m,&c1,&c2);
	for(int i=0;i<n;++i)scanf("%d",&weight[i]);
	memset(G,INF,sizeof(G));
	for(int i=0;i<m;++i){
		scanf("%d%d",&u,&v);
		scanf("%d",&G[u][v]);
		G[v][u]=G[u][v];
	}
	
	Dijkstra(c1);
	cout<<num[c2]<<" "<<w[c2]<<endl;
	return 0;
}