单源最短路径的算法，存在 相等长度的路径，再做额外判断即可。

#include <bits/stdc++.h>
#define MAX 1000000
using namespace std;
int n,m,a,b;
int G[501][501];
int resc[501];
int num = 1;
int dis[501];
int visited[501] = {0};
int totalresc[501];
int pre[501];
int FindMin(){
	int MinDist = MAX,MinV = -1;
	for(int i = 0; i < n; i++)
		if(!visited[i] && dis[i] < MinDist){
			MinDist = dis[i];
			MinV = i;
		}
	return MinV;
}
void dijkstra(){
	for(int i = 0; i <n; i++){
		dis[i] = G[a][i];
		if(G[a][i] != MAX){
			totalresc[i] = resc[a] + resc[i];
			pre[i] = 1;
		}
	 	else{
	 		totalresc[i] = resc[i];
	 		pre[i] = 0;
		}
	}
	dis[a] = 0;
	pre[a] = 1;
	visited[a] = 1;
	totalresc[a] = resc[a];
	while(1){
		int nod = FindMin();
		if(nod == -1) break;
		visited[nod] = 1;
		for(int i = 0; i < n; i++){
			if(!visited[i])
				if(dis[nod] + G[nod][i] < dis[i]){
					dis[i] = dis[nod] + G[nod][i];
					totalresc[i] = totalresc[nod] + resc[i];
					pre[i] = pre[nod];
					
				}
				else if(dis[nod] + G[nod][i] == dis[i]){
					pre[i] = pre[i] + pre[nod];
					if(totalresc[nod] + resc[i] > totalresc[i])
					totalresc[i] = totalresc[nod] + resc[i];
			    }
		}
	}
}
int main(){
	//freopen("1.txt","r",stdin);
	int x,y,l;
	cin >> n >> m >> a >> b;
	for(int i = 0; i < n; i++)
		for(int j = 0; j < n; j++) G[i][j] = MAX;
	for(int i = 0; i < n; i++) cin >> resc[i];
	while(m--){
		cin >> x >> y >> l;
		G[x][y] = l;
		G[y][x] = l;
	}
	dijkstra();
	cout << pre[b] << " " << totalresc[b];
}