// https://dashoj.com/p/80
#include <bits/stdc++.h>

#define N 50
using namespace std;
typedef long long ll;
int n, g[N][N], e; // 
bool gb[N];
ll ans = 0x3f3f3f3f;

void dfs(int x, int a, ll an) {
	if (a == e) {
		ans = min(an + g[x][n], ans);
		return;
	}
	for (int i = 2; i <= e; i++) {
		if (i == x) continue;
		if (!gb[i]) continue;
		gb[i] = false;
		if (an + g[x][i] < ans) dfs(i, a + 1, an + g[x][i]);
		gb[i] = true;
	}
}

int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr); //
	fill(&gb[0], &gb[0] + N, true); //
	cin >> n; //
	e = n - 1; //
	gb[1] = false;
	for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) cin >> g[i][j]; //
	dfs(1, 1, 0); //
	cout << ans << endl; //
	return 0;
}

开O2优化可以用dfsAC

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N = 21;

int n;
int d[N][N];
bool st[N];
LL ans = 0x3f3f3f3f;

void dfs(int u, int cnt, LL cost)// 当前所在岛屿， 总共经过了多少个岛屿， 经过cnt个岛屿的花费
{
	if(cost > ans) return;
	if (cnt == n && u == n)
	{
		ans = min(ans, cost);
		return;
	}

	for (int i = 1; i <= n; i++)
	{
		if (i == u) continue;
		if (st[i]) continue;
		if (d[u][i] >= ans) continue;

		st[i] = true;
		dfs(i, cnt + 1, cost + d[u][i]);
		st[i] = false;

	}
}

void solve()
{
	cin >> n;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			cin >> d[i][j];
			
	st[1] = true;
	dfs(1, 1, 0);
}

int main()
{
	solve();
	cout << ans << endl;
	return 0;
}

#include<bits/stdc++.h> using namespace std; // typedef long long ll; const int N=21; int n; int d[N][N]; bool st[N]; int ans=0x3f3f3f3f; void dfs(int u,int cnt,int cost) { if(cost>ans) return; if(cntn&&un) { ans=min(ans,cost); return; } for(int i=1;i<=n;i++) { if(i==u) continue; if(st[i]) continue; if(d[u][i]>ans) continue; st[i]=true; dfs(i,cnt+1,cost+d[u][i]); st[i]=false;; } } int main() { cin>>n; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { cin>>d[i][j]; } } st[1]=true; dfs(1,1,0); cout<<ans; return 0; }

#include<bits/stdc++.h>
using namespace std;
int n;int mp[20][20];
bool vis[20];
using ll=long long;
ll ans=1e9+5;

void dfs(int x,ll ann){//x是到了哪座岛出发，ann是花费
    if(x==n){
        for(int i=1;i<n;i++){
            if(!vis[i])return;
        }
        ans=min(ans,ann);
        return;
    }
    if(ann>ans){
        return;
    }
    for(int j=1;j<=n;j++){
        if(x==j) continue;
        if(vis[j]) continue;
        vis[j]=true;
        dfs(j,ann+mp[x][j]);
        vis[j]=false;
    }

}
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            cin>>mp[i][j];
        }
    }
    vis[1]=true;
    dfs(1,0);
    cout<<ans<<endl;
    return 0;
}

使用dfs + 剪枝，最多可过70%数据

#include <bits/stdc++.h>
using namespace std;
const int N = 20;
const int INF = 1e9;
int mp[N][N];
int vis[N];
int n;
int mn = INF;
int mem[N][N][N];
int dfs(int x, int sum, int cnt) {
    if (cnt == n && x == n) {
        mn = min(mn, sum);
        return mn;
    }
    if (cnt == n || sum >= mn) {
        return INF;
    }
    for (int i = 1; i <= n; ++i) {
        if (!vis[i]) {
            vis[i] = 1;
            return dfs(i, sum + mp[x][i], cnt + 1);
            vis[i] = 0;
        }
    }
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            scanf("%d", &mp[i][j]);
        }
    }
    vis[1] = 1;
    printf("%d\n", dfs(1,0,1));
    return 0;
}

正解DP

#include <bits/stdc++.h>
using namespace std;
const int N = 20;
const int INF = 1e9;
int mp[N][N];
int dp[1 << N][N];
int n;

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            scanf("%d", &mp[i][j]);
        }
    }
    for (int i = 0; i < (1 << n); ++i) {
        for (int j = 0; j < n; ++j) {
            dp[i][j] = INF;
        }
    }
    dp[1][0] = 0;
    for (int i = 1; i < (1 << n); ++i) {
        for (int j = 0; j < n; ++j) {
            if ((i >> j) & 1) {
                for (int k = 0; k < n; ++k) {
                    if ((i >> k) & 1 && j != k) {
                        dp[i][j] = min(dp[i][j], dp[i ^ (1 << j)][k] + mp[k][j]);
                    }
                }
            }
        }
    }
    printf("%d\n", dp[(1 << n) - 1][n - 1]);
    return 0;
}