HDU 1565 方格取数(1)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1565

思路:跟上一篇 POJ 的状压dp很类似在这里就不过多阐述了

#include<iostream>
#include<algorithm>
#include<cstring>
#define M 20000
using namespace std;
typedef long long LL;
LL n, pass[M], now[22][M], mp[25][25];
LL dp[22][M];//这里之所以不开1<<n 是因为虽然1<<n很大，但是其中有效状态比较小所以只开M个就足够了
bool check(int x, int y) {
    return (pass[x] & pass[y]);
}
int main()
{
    ios::sync_with_stdio(false);
    while (cin >> n) {
        if (n == 0) {
            cout << "0" << endl;
            continue;
        }
        memset(pass, 0, sizeof(pass));
        memset(now, 0, sizeof(now));
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                cin >> mp[i][j];
        int cnt = 0;
        for (int i = 0; i < (1 << n); i++)
            if (!(i&(i << 1)))//将可行状态保存下来
                pass[cnt++] = i;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < cnt; j++) {
                int t = n;
                for (int k = 1; k < (1 << n); k<<=1) {
                    if ((pass[j]&k))
                        now[i][j] += mp[i][t];//now[i][j]代表第i行第j种状态的获取值
                    t--;
                }
            }
        }
        for (int i = 0; i < cnt; i++)
            dp[1][i] = now[1][i];
        for (int i = 2; i <= n; i++) {
            for (int j = 0; j < cnt; j++) {
                for (int k = 0; k < cnt; k++) {
                    if (!check(j, k))
                        dp[i][j] = max(dp[i][j], dp[i - 1][k] + now[i][j]);
                }
            }
        }
        LL ans = 0;
        for (int i = 0; i < cnt; i++)
            ans = max(dp[n][i], ans);
        cout << ans << endl;
    }
    return 0;
}