POJ1185+状态压缩DP

经典的状态压缩DP。

dp[i][j][k] = max( dp,dp[i-1][k][k2] );

k是前一行的状态，k2是前二行的状态。

/*
dp[i][j][k] = max( dp,dp[i-1][k][k2] );
k是前一行的状态，k2是前二行的状态。
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<math.h>
#include<map>
using namespace std;
const int maxn = 105;
const int maxm = 12;
const int N = 65;
int mat[ maxn ],mm[ maxn ][ maxm ];
int dp[ maxn ][ N ][ N ];
int state[ maxn ],ones_state[ maxn ];

int GetSum( int x ){
	int sum = 0;
	while( x ){
		if( x&1 )
			sum++;
		x>>=1;
	}
	return sum;
}		
	
int init_state( int n,int m ){
	int M = (1<<m);
	int cnt =0;
	memset( state,0,sizeof( state ) );
	memset( ones_state,0,sizeof( ones_state ) );
	for( int i=0;i<M;i++ ){
		if( (i&(i<<1))==0&&(i&(i<<2))==0&&(i&(i>>1))==0&&(i&(i>>2))==0 ){
			state[ cnt ] = i;
			ones_state[ cnt++ ] = GetSum( i );
		}
	}
	//printf("cnt=%d\n",cnt);
	return cnt;
}

void DP( int cnt,int n,int m ){
	memset( dp,-1,sizeof( dp ) );
	for( int i=0;i<cnt;i++ ){
		if(( mat[0]&state[i] )==0)
			dp[ 0 ][ i ][ 0 ] = ones_state[ i ];
	}
	for( int i=1;i<n;i++ ){
		for( int j=0;j<cnt;j++ ){
			if( (mat[i]&state[j])==0 ){
				for( int k=0;k<cnt;k++ ){
					if( (state[j]&state[k])==0 ){
						for( int k2=0;k2<cnt;k2++ ){
							if( dp[i-1][k][k2]==-1 ) continue;
							if( (state[j]&state[k2])==0&&(state[k]&state[k2])==0 ){
								dp[i][j][k] = max( dp[i][j][k],dp[i-1][k][k2]+ones_state[j] );
							}
						}
					}
				}
			}
		}
	}
	return ;
}

int main(){
	int n,m;
	while( scanf("%d%d",&n,&m)==2 ){
		int cnt = init_state( n,m );
		char s[ maxm ];
		for( int i=0;i<n;i++ ){
			scanf("%s",s);
			for( int j=0;j<m;j++ ){
				if( s[j]=='P' )
					mm[i][j] = 1;
				else
					mm[i][j] = 0;
			}
		}
		memset( mat,0,sizeof( mat ) );
		for( int i=0;i<n;i++ ){
			for( int j=0;j<m;j++ ){
				if( mm[i][j]==0 )
					mat[i] |= (1<<j);
			}
		}
		DP( cnt,n,m );
		int ans = 0;
		for( int i=0;i<cnt;i++ ){
			for( int j=0;j<cnt;j++ ){
				ans = max( ans,dp[n-1][i][j] );
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}