查看题目:1003 Emergency
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 <iostream>
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;
}
原文:https://www.cnblogs.com/qujunhui/p/10884382.html