power oj 2480 放积木[二进制状压DP]

题目链接【https://www.oj.swust.edu.cn/problem/show/2480】

题意：中文题目。

题解：二进制状态转移+坏点判断。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn =  << ;
LL dp[][maxn];
int mp[];
char s[];
int n, m;
int cur;
void DFS(int r, int pos, int nu, int z, int cnt)
{
    if(pos == m)
    {
        dp[cur][nu] = max(dp[cur][nu], dp[cur ^ ][z] + cnt);
        return ;
    }
    if(mp[r] & ( << pos)) //坏点
    {
        DFS(r, pos + , nu, z, cnt);
        return ;
    }
    if(!(z & ( << pos)) && !(mp[r - ] & ( << pos)))
        DFS(r, pos + , nu | ( << pos), z, cnt + );
    else
    {
        if( pos && !(nu & ( << (pos - ))) && !(mp[r] & ( << (pos - ))) )
            DFS(r, pos + , nu | ( << pos) | ( << (pos - )), z, cnt + );
        DFS(r, pos + , nu, z, cnt);
    }
}
int main ()
{
    while(~scanf("%d%d", &n, &m))
    {
        memset(mp, , sizeof(mp));
        for(int i = ; i <= n; i++)
        {
            scanf("%s", s);
            for(int j = ; j < m; j++)
                if(s[j] == '*') mp[i] |=  << j;
        }
        cur = ;
        memset(dp[], -, sizeof(dp[]));
        int mask = ( << m) - ;
        dp[][mask] = ;
        for(int i = ; i <= n; i++)
        {
            cur ^= ;
            memset(dp[cur], -, sizeof(dp[cur]));
            for(int j = ; j <= mask; j++)
                if(dp[cur ^ ][j] >= )
                    DFS(i, , , j, );
        }
        LL ans = ;
        for(int i = ; i <= mask; i++)
            ans = max(ans, dp[cur][i]);
        printf("%lld\n", ans);
    }
    return ;
}