旅行商问题用回溯法以及排列空间树解决，请给我具体C++实现代码

好的，以下是旅行商问题的回溯法和排列空间树的C++实现代码：

回溯法：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 20;

int n, ans = INF;
int map[MAXN][MAXN];
bool vis[MAXN];

void dfs(int cur, int cost, int cnt) {
    if (cnt == n && map[cur][0] != INF) {
        ans = min(ans, cost + map[cur][0]);
        return;
    }
    if (cost >= ans) return;
    for (int i = 1; i <= n; i++) {
        if (!vis[i] && map[cur][i] != INF) {
            vis[i] = true;
            dfs(i, cost + map[cur][i], cnt + 1);
            vis[i] = false;
        }
    }
}

int main() {
    memset(vis, false, sizeof(vis));
    memset(map, INF, sizeof(map));
    cin >> n;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> map[i][j];
        }
    }
    vis[0] = true;
    dfs(0, 0, 0);
    cout << ans << endl;
    return 0;
}

排列空间树：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 20;

int n, ans = INF;
int map[MAXN][MAXN];
bool vis[MAXN];

void dfs(int cur, int cost, int cnt) {
    if (cnt == n && map[cur][0] != INF) {
        ans = min(ans, cost + map[cur][0]);
        return;
    }
    if (cost >= ans) return;
    for (int i = 1; i <= n; i++) {
        if (!vis[i] && map[cur][i] != INF) {
            vis[i] = true;
            dfs(i, cost + map[cur][i], cnt + 1);
            vis[i] = false;
        }
    }
}

int main() {
    memset(vis, false, sizeof(vis));
    memset(map, INF, sizeof(map));
    cin >> n;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> map[i][j];
        }
    }
    vis[0] = true;
    dfs(0, 0, 0);
    cout << ans << endl;
    return 0;
}