感觉这道题目的数据比较水啊，程序的时间复杂度为1711^2*20竟然也可以过掉。。。。其他的就是状压了啊，注意需要滚动一下啊。。。。

方格取数(1)

3
75 15 21 
75 15 28 
34 70 5 

188

#include <set>
#include <map>
#include <queue>
#include <math.h>
#include <vector>
#include <string>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>


using namespace std;

const int maxn = 22;
int dp[2][1<<21];
int mp[maxn][maxn];
int num[1<<22];

int judge(int x)
{
    for(int i = 0; i < 20; i++) if((x&(1<<i)) && (x&(1<<(i+1)))) return 1;
    return 0;
}

int main()
{
    int n;
    int pp = 0;
    for(int i = 0; i < (1<<20); i++) if(!judge(i)) num[pp++] = i;
    cout<<pp<<endl;
    while(~scanf("%d", &n))
    {
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++) scanf("%d",&mp[i][j]);
        memset(dp, 0, sizeof(dp));
        int flag;
        for(int p = 0; p < pp; p++)
        {
            int i = num[p];
            flag = 0;
            for(int j = 0; j < n; j++)
            {
                if((i&(1<<j)) && (i&(1<<(j+1))))
                {
                    flag = 1;
                    break;
                }
                if(i&(1<<j)) dp[0][i] += mp[0][j];
            }
            if(flag) dp[0][i] = 0;
        }

        for(int i = 1; i < n; i++)
        {
            for(int p = 0; p < pp && num[p] < (1<<n); p++)
            {
                int j = num[p];
                dp[1][j] = 0;
                int flag = 0;
                for(int k = 0; k < n-1; k++)
                {
                    if((j&(1<<k)) && (j&(1<<(k+1))))
                    {
                        flag = 1;
                        break;
                    }
                }
                if(flag) continue;
                for(int kk = 0; kk < n; kk++) if(j&(1<<kk)) dp[1][j] += mp[i][kk];
                int sum = 0;
                for(int tt = 0; tt < pp && num[tt] < (1<<n); tt++)
                {
                    int k = num[tt];
                    if(!dp[0][k] && k > 0) continue;
                    if(j&k) continue;
                    sum = max(sum, dp[0][k]);
                }
                dp[1][j] += sum;
            }
            for(int j = 0; j < pp && num[j] < (1<<n); j++) dp[0][num[j]] = dp[1][num[j]];
        }
        int Max = 0;
        for(int i = 0; i < pp && num[i] < (1<<n); i++) Max = max(dp[0][num[i]], Max);
        printf("%d\n",Max);
    }
    return 0;
}

/*
20
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
*/

HDU 1565 方格取数(1)(状压dp)

原文：http://blog.csdn.net/xu12110501127/article/details/40053609