poj 1185

http://poj.org/problem?id=1185

/*
Problem: 1185        User: ruan123
Memory: 2128K        Time: 266MS
Language: G++        Result: Accepted
说下dp方程：dp[i][j][k] 为第i行状态为Sj，第i-1行状态为Sk时前i行最多能放置的炮数。 
*/


#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int dp[105][65][65];
int s[105],map[105],cnt,c[100];

void init(int m)
{
    m = 1<<m;
    for(int i = 0; i < m; ++ i){
        if(!(i&(i<<1)) && !(i&(i<<2))){
            int sum = 0;
            int k = i;
            while(k){ sum += k&1; k /= 2; }
            c[cnt] = sum;
            s[cnt++] = i;
        }
    }
}

int main()
{
    int n,m,ans = 0;
    char str[100];
    cnt = 0;
    
    scanf("%d%d",&n,&m);
    for(int i = 0; i < n; ++ i){
        scanf("%s",str);
        for(int j = 0; str[j]; ++j )
            if(str[j] == 'H') map[i] += (1 << j);
    }
    
    init(m);
    memset(dp,-1,sizeof(dp));
    for(int i = 0; i < cnt; ++ i)
        if(!(map[0]&s[i])) dp[0][i][0] = c[i];
        
    for(int i = 1; i < n; ++ i){
        for(int j = 0; j < cnt; ++ j){
            if(s[j]&map[i])continue;//有冲突 
            for(int k = 0; k < cnt; ++ k){
                if(s[j]&s[k])continue;//有冲突
                for(int l = 0; l < cnt; ++ l){
                    if(s[j]&s[l])continue;//有冲突
                    if(dp[i-1][k][l]<0)continue;//没有值 
                    dp[i][j][k] = dp[i][j][k] > dp[i-1][k][l] + c[j] ? dp[i][j][k] : dp[i-1][k][l] + c[j];
                }
            }
        }
    }
    
    for(int i = 0; i < cnt; ++ i)
        for(int j = 0; j < cnt; ++ j){
            ans = dp[n-1][i][j] > ans? dp[n-1][i][j] : ans;
        }
    printf("%d\n",ans);
    return 0;
}