源哥每日一题第十四弹 hdu 1565 还是状压dp

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

题意：自己念

分析：有了上一题的基础，这题的状态方程也好想：dp[i][s]表示第i行第s个状态取得的最大值。

　　可以预处理出每一行每一种状态所能获得的值，方程就是dp[i][j] = max(dp[i][j],dp[i-1][k]+sum[i][j]);

　　要注意的就是判断相邻的情况：两个状态，如果a&b！=0则表示它们有相重叠的区域。

#include <bits/stdc++.h>

using namespace std;

void read(){
#ifndef ONLINE_JUDGE
    freopen("D:\\fengyu\\Jiang_C\\.vscode\\in.txt","r",stdin);
	freopen("D:\\fengyu\\Jiang_C\\.vscode\\out.txt","w",stdout);
#endif
}
#define clr(a) memset(a,0,sizeof(a))
#define ll long long
ll a[25][25],s[20005];
ll dp[25][20005];
ll sum[25][20005];
int main() { read();
    int n;
    int num = 0;
    for (int i = 1; i< 1<<20; i++) {
        if(i&i<<1)
            continue;
        s[num++]=i;
    }
    while (cin >> n) {
        clr(dp);
        clr(sum);
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                cin >> a[i][j];
            }
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 0; s[j] < 1<<n; j++) {
                int t = s[j];
                for (int k = 1; k <= n; k++) {
                    if(t&1) sum[i][j]+=a[i][k];
                    t>>=1;
                }
            }
        }
        for (int j = 0; s[j] < 1<<n; j++) {
            dp[1][j] = sum[1][j];
        }

        for (int i = 2; i <= n; i++) {
            for(int j = 0; s[j] < 1<<n; j++) {
                for(int k = 0; s[k] < 1<<n; k++) {
                    if (s[k] & s[j]) continue;
                    dp[i][j] = max(dp[i][j],dp[i-1][k]+sum[i][j]);  

                }
            }
        }
        ll ans = 0;
        for (int j = 0; s[j] < 1<<n; j++) {
            ans = max(ans,dp[n][j]);
        }
        cout << ans << endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/fengyuzhicheng/p/9157960.html