CH0103 最短Hamilton路径 状态压缩

//CH0103 2020/10/07 23:21
#include <bits/stdc++.h>
using namespace std;

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

int n, mp[N][N], val[N][1 << 20];

struct Node {
    int cur, sit;
};

queue <Node> q;

int main () {
    cin >> n;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            cin >> mp[i][j];
        }
    }
    memset (val, 0x3f, sizeof (val));
    val[0][1] = 0;
    q.push ((Node) {0, 1}); // 从第 0 位开始
    while (!q.empty ()) {
        Node _node = q.front ();
        q.pop ();
        // cout << "cur = " << _node.cur << ", sit = " << _node.sit << ", val = " << val[_node.cur][_node.sit] << endl;
        for (int i = 0; i < n; ++i) {
            if (((_node.sit >> i) & 1) == 0) {
                int _nextval = val[_node.cur][_node.sit] + mp[_node.cur][i];
                int _nextsit = _node.sit + (1 << i);
                // cout << "nextval = " << _nextval << ", nextsit = " << _nextsit << endl;
                if (val[i][_nextsit] > _nextval) {
                    if (val[i][_nextsit] == INF) {
                        q.push ((Node){i, _nextsit});
                    }
                    val[i][_nextsit] = _nextval;
                }
            }
        }
    }
    cout << val[n - 1][(1 << n) - 1] << endl;
}

基础的状压dp，每一层的点不会相互重复所以采取bfs。