这题可以使用dfs的方法结合stack的使用进行求回路
#include<iostream>
#include<stack>
#include<cstring>
#pragma(disable:4996)
using namespace std;
const int N = 20;
int first, click, times, a[N][N], vis[N];
//first表示出发点
//click用来标记以经走过多少个城市
//times计算可走的路线数目
//a为邻接矩阵
//vis用来标记城市是否走过
void dfs(stack<int> &S, int k) {
if (click == N + 1 && first == S.top()) {//走完所有的城市且回到出发地,输出当前的路线
int b[N + 1];
for (int i = N; i >= 0; i--) {
b[i] = S.top();
S.pop();
}
for (int i = 0; i <= N; i++)
S.push(b[i]);
cout << times++ << ": ";
for (int i = 0; i <= N; i++) {
cout << " " << b[i] + 1;
}
cout << endl;
}
for (int i = 0; i < N; i++) {
if (i != k && vis[i] == 0 && a[k][i] == 1) {//寻找相邻且当前路线没有走过的城市
S.push(i);
vis[i] = 1;
click++;
dfs(S, i);
S.pop();
vis[i] = 0;
click--;
}
}
}
int main() {
int x;
stack<int> S;
memset(a, 0, sizeof(a));
memset(vis, 0, sizeof(vis));
for (int i = 0; i < N; i++)
for (int j = 0; j < 3; j++) {
scanf("%d", &x);
a[i][x - 1] = 1;
a[x - 1][i] = 1;
}
while (scanf("%d", &x) != 0 && x != 0) {
click = 1;
S.push(x - 1);
first = x - 1;
times = 1;
dfs(S, x - 1);
}
return 0;
}