枚举子集+预处理优化dp+贪心视角转化成可做dp

https://www.acwing.com/problem/content/531/

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 12, M = 1 << N, INF = 0x3f3f3f3f;

int n, m;
int d[N][N], g[N][M];
int f[M][N];

int main()
{
    scanf("%d%d", &n, &m);

    memset(d, 0x3f, sizeof d);
    for (int i = 0; i < n; i ++ ) d[i][i] = 0;
    while (m -- )
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        a --, b -- ;
        d[a][b] = d[b][a] = min(d[a][b], c);
    }

    memset(g, 0x3f, sizeof g);
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < 1 << n; j ++ )
            for (int k = 0; k < n; k ++ )
                if (j >> k & 1)
                    g[i][j] = min(g[i][j], d[i][k]);
                    //预处理加速dp转移

    memset(f, 0x3f, sizeof f);
    for (int i = 0; i < n; i ++ ) f[1 << i][0] = 0;//利用题目条件：初始道路免费
    
    
    for (int i = 1; i < 1 << n; i ++ )
        for (int j = i - 1 & i; j; j = j - 1 & i)  // 枚举i的子集
        {
            int r = i ^ j, cost = 0;
            for (int k = 0; k < n; k ++ )
                if (j >> k & 1)
                {
                    cost += g[k][r];
                    if (cost >= INF) break;
                }
            if (cost >= INF) continue;
            for (int k = 1; k < n; k ++ )
                f[i][k] = min(f[i][k], f[r][k - 1] + cost * k);
        }

    int res = INF;
    for (int i = 0; i < n; i ++ )
        res = min(res, f[(1 << n) - 1][i]);
    printf("%d\n", res);

    return 0;
}

__EOF__

本文作者：爱飞鱼
本文链接：https://www.cnblogs.com/mathiter/p/18088584.html
关于博主：评论和私信会在第一时间回复。或者直接私信我。
版权声明：本博客所有文章除特别声明外，均采用 BY-NC-SA 许可协议。转载请注明出处！
声援博主：如果您觉得文章对您有帮助，可以点击文章右下角【推荐】一下。您的鼓励是博主的最大动力！