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声明：程序源码并非完美，提交后结果为“部分正确”。因博主看到这个题目网上的答案较少，所以将未完善的代码发出来供大家参考！

本程序采用深度优先搜索算法（DFS），由于递归的调用方式，占用内存过大导致程序提交错误。

#include "pch.h"#include 
     
      using namespace std;const int MAXV = 50;const int INF = 99999;struct Graph {    int Vex[MAXV];    int arcs[MAXV][MAXV];    int vexnum, arcsnum;}G;bool visited[MAXV];int path[MAXV];int length = 0;int min_length = INF;int num = 0;int team = 0;int maxteam = 0;void DFS(Graph G, int c1, int c2) {    team += G.Vex[c1];    if (c1 == c2) {
    //走到终点了        //length += G.arcs[c1][c2];        if (length < min_length) {            num = 1;            min_length = length;            maxteam = team;        }        else if (length == min_length) {            num++;            if (team > maxteam)maxteam = team;        }        return;    }    visited[c1] = true;        for (int i = 0; i < G.vexnum; i++) {        if (G.arcs[c1][i] != INF && visited[i] == false) {
    //访问v的邻接点            length += G.arcs[c1][i];            //team += G.Vex[i];            DFS(G,i,c2);            length -= G.arcs[c1][i];            //team -= G.Vex[i];        }    }    team -= G.Vex[c1];    return;}int main() {    for (int i = 0; i < MAXV; i++) {        for (int j = 0; j < MAXV; j++) {            G.arcs[i][j] = INF;        }    }    int c1, c2;    cin >> G.vexnum >> G.arcsnum >> c1 >> c2;        for (int i = 0; i < G.vexnum; i++) {
    //城市救援队数量        cin >> G.Vex[i];    }    int temp1, temp2, temp3;    for (int i = 0; i < G.arcsnum; i++) {        cin >> temp1 >> temp2 >> temp3;        G.arcs[temp1][temp2] = temp3;    }    DFS(G, c1, c2);    cout << num <<' '<< maxteam << endl;}