1. 用迪杰斯特拉算法求最短路径
2. 用before的vector数组记录当前的结点的所有上一个结点，通过这种方式记录下一个以出发结点和终点为起点和终点，用最短路径构成的拓扑图（所有以起点和终点的最短路径长度相等）
3. 用深度优先算法遍历各个拓扑图并获得各条路上的救火队数目，并计算最短路径的条数。

#include<cstdio>
#include<queue>

using namespace std;

const int maxn = 505;
const int inf = 1000000000;

int teamNum[maxn];
int road[maxn][maxn];
int visit[maxn], cost[maxn];
vector<int> before[maxn];

vector<int> q, remenber;
int maxTeam = 0, projectNum = 0;
int start;

void DFS(int aim) {
    if (aim == start) {
        projectNum++;
        int totalTeam = 0;
        //printf("\n");  //
        for (int i = 0; i < q.size(); i++) {
            totalTeam += teamNum[q[i]];
            // printf("%d ", q[i]);
        }
        if (totalTeam > maxTeam) {
            maxTeam = totalTeam;
            remenber = q;
        }
        return;
    }
    for (int i = 0; i < before[aim].size(); i++) {
        q.push_back(before[aim][i]);
        DFS(before[aim][i]);
        q.pop_back();
    }
    return;
}

int main() {


    int N, M, aim;
    scanf("%d %d %d %d", &N, &M, &start, &aim);
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (i == j) road[i][i] = 0;
            else road[i][j] = inf;
        }
    }
    for (int i = 0; i < N; i++) {
        visit[i] = inf;
        cost[i] = inf;
    }
    for (int i = 0; i < N; i++) {
        scanf("%d", &teamNum[i]);
    }
    int c1, c2, length;
    for (int i = 0; i < M; i++) {
        scanf("%d %d %d", &c1, &c2, &length);
        road[c1][c2] = length;
        road[c2][c1] = length;
    }
    visit[start] = 0;
    int smaller = start, m = inf;
    bool done = false;
    for (int k = 0; k < N; k++) {
        for (int i = 0; i < N; i++) {
            if (i == smaller) continue;
            if (road[smaller][i] + visit[smaller] < cost[i]) {
                cost[i] = road[smaller][i] + visit[smaller];
                before[i].clear();
                before[i].push_back(smaller);
            }
            else if (road[smaller][i] + visit[smaller] == cost[i]) {
                before[i].push_back(smaller);
            }
        }
        if (done == true) break;
        smaller = -1, m = inf;
        for (int i = 0; i < N; i++) {
            if (visit[i] == inf && cost[i] < m) {
                m = cost[i];
                smaller = i;
            }
        }
        visit[smaller] = cost[smaller];
        if (smaller == aim) done = true;
    }
    q.push_back(aim);
    DFS(aim);

    printf("%d %d", projectNum, maxTeam);
    return 0;

}