vijo 1456最小总代价

题意：中文题。。。

题解：状态比较多，可以说是状压的基础题吧，我们定义dp[i][j],j为一个二进制数，每位0表示接触过该物品，1表示没有接触过;j表示当前物品在谁手上。递推的顺序注意一下就好

ac代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

#define MAXN 20
#define INF 0x3f3f3f3f

using namespace std;

int n;
int edges[MAXN][MAXN];
int dp[65546][MAXN];

int min(int a, int b)
{
    if (a == -1) return b;
    if (b == -1) return a;
    return a < b ? a : b;
}

int main()
{
    scanf("%d", &n);
    int t;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            scanf("%d", &edges[i][j]);
        }
    }
    memset(dp, -1, sizeof(dp));
    for (int i = 0; i < n; i++)
    {
        dp[1 << i][i] = 0;
    }
    int ans = -1;
    for (int i = 0; i < 1 << n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            if (dp[i][j] != -1)
            {
                for (int k = 0; k < n; k++)
                {
                    if (!(i & (1 << k)))
                    {
                        dp[i | (1 << k)][k] = min(dp[i | (1 << k)][k], dp[i][j] + edges[j][k]);
                        if ((i | (1 << k)) == (1 << n) - 1) ans = min(ans, dp[i | (1 << k)][k]);
                    }
                }
            }
        }
    }
    if (ans != -1)
        printf("%d
", ans);
    else printf("0
");

    return 0;
}