单源最短路径
使用Dijkstra求解

#include<cstdio>
#include<algorithm>
using namespace std;
int N,M,S,D;
const int maxn=505;
const int INF=1e9;
int G[maxn][maxn];
int C[maxn][maxn];
int vis[maxn]={0};
int d[maxn];
int costs[maxn];
void Dijkstra(int s){
	fill(d,d+maxn,INF);
	fill(costs,costs+maxn,INF);
	d[s]=0;
	costs[s]=0;
	for(int i=0;i<N;i++){
		int Min = INF,u = -1;
		for(int j=0;j<N;j++){
			if(vis[j]==0 && d[j] < Min){
			u=j;
			Min=d[j];
			}
		}

	if(u ==  -1) return;
	vis[u]=1;
	for(int v=0;v<N	;v++){ //以u为父节点，延伸出来的边 
			if(vis[v]==0 && G[u][v]!=INF ){
				if( d[u] +G[u][v] < d[v]){
					d[v]= G[u][v] + d[u];
					costs[v]=costs[u]+C[u][v];
				}else if(d[u] +G[u][v] == d[v] && costs[u] +C[u][v] < costs[v]){
					costs[v]=costs[u]+C[u][v];
				}
				
			}
	}
  }
}
int main(){
	int u,v,length,cost;
	scanf("%d%d%d%d",&N,&M,&S,&D);
	fill(G[0],G[0]+maxn*maxn,INF);
	for(int i=0;i<M;i++){
		scanf("%d%d%d%d",&u,&v,&length,&cost);
		G[u][v]=length;
		G[v][u]=length;
		C[u][v]=cost;
		C[v][u]=cost;
	}
	Dijkstra(S);
	printf("%d %d\n",d[D],costs[D]);
	return 0;
}