acwing 91. 最短Hamilton路径(哈密尔顿)

传送门
状压dp

#include

using namespace std;
const int N =20, M = 1<<20;

int f[M][N], d[N][N], n;
///f[i][j]代表经过状态i，终点在j的点，i代表状态，转成二进制就代表经过那些点
int main()
{
    scanf("%d", &n);
    for(int i = 0; i < n; ++i)
    for(int j = 0; j < n; ++j)
    scanf("%d", &d[i][j]);
    memset(f, 0x3f, sizeof(f));
    f[1][0] = 0;
    
    for(int i = 0; i < 1<<n; ++i)
    {
        for(int j = 0; j < n; ++j) 
            if(i>>j&1)
            for(int k = 0; k < n; ++k)
            if(i-(1<<j)>>k&1)
            f[i][j] = min(f[i][j], f[i-(1<<j)][k]+d[k][j]);
    }
    printf("%d\n", f[(1<<n)-1][n-1]);
    return 0;
}