原题链接

• 题意：给\(n <= 24\)个点，然后给出起点，从起点出发，一个人一次只能拿一个或者两个东西，然后放回起点，求最短路径长度，并输出方案数。
• 题解：应该是状压dp,但是并没有想清楚，一开始就写了一个假暴力。\(dp_s\)代表了拿这些物品的最短路径长度，然后应该是\(dp_0 = 0\)显然，然后应当从已知状态传递到未知状态，设状态为\(s\), 所以枚举 \(i, j\)，当 \(s\) 中没有 \(i\) 和 \(j\) 的时候，就可以通过 \(s\) 来传递到 \(s | 1 << (i - 1) | 1 << (j - 1)\) 的后面状态，然后即 \(dp_{s | 1 << (i - 1) | 1 << (j - 1)} = \min (dp_{s} + dis(i, j) + dis(i, 0) + dis(j, 0))\).
• 代码：

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>

using namespace std;
typedef long long ll;
const int N = 33;

int mp[N][N];
int sx, sy;
struct node {
    int x, y;
}a[N];
int dp[1 << 25];
int pre[1 << 25];
int dis(int i, int j) {
    return (a[i].x - a[j].x ) * (a[i].x - a[j].x ) + (a[i].y - a[j].y ) * (a[i].y - a[j].y );
}
vector<int>ans;
void solve() {
    cin >> sx >> sy;
    int n;cin >> n;
    for (int i = 1; i <= n; i ++) cin >> a[i].x >> a[i].y;
    a[0] = {sx, sy};
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (i==j)continue;
            mp[i][j] = (a[i].x- a[j].x) * (a[i].x - a[j].x) + (a[i].y - a[j].y) * (a[i].y - a[j].y);
        }
    }
    int t = 0;
    memset(dp, 0x3f, sizeof dp);
    memset(pre, -1, sizeof pre);
    dp[0] = 0;
    int inf = 0x3f3f3f3f3f3f3f3f;
    for (int s = 0; s < 1 << n; s++) {
        if (dp[s] == inf)continue;
        for (int i = 1; i <= n; i++) {
            if (  s & 1 << (i-1))continue;
            for (int j = 1; j <= n; j++) {
                if ( s & 1 << (j - 1)) continue;
                if (dp[s | (1 << (i-1)) | (1 << (j-1))] > dp[s] + dis(j, 0) + dis(i, j) + dis(i, 0)) {
                    dp[s | (1 << (i-1)) | (1 << (j-1))] = dp[s] + dis(j, 0) + dis(i, j) + dis(i, 0);
                    pre[s | (1 << (i-1)) | (1 << (j-1))] = s;
                }
            }
            break;
        }
    }
    int now = (1 << n) - 1;
    ans.push_back(0);
    while (now != -1) {
        int s = pre[now];
        if (s == -1)s = 0;
        bool f = 1;
        for (int i = 1; i <= n; i++) {
            if ( ((s >> (i-1) ) & 1) != ((now >> (i-1) ) & 1)) {
                ans.push_back(i);
                f = 0;
            }
        }
        ans.push_back(0);
        now = s;
        if (s == 0)break;
    }
    cout << dp[ (1 << n)  -1] << endl;
    for (auto iter: ans) cout << iter << " ";
}
signed main() {
    int t = 1;
    while (t--) solve();
    return 0;
}