POJ 1185 炮兵阵地（状压DP）题解

思路：和上一篇思路一样，但是这里要求最大能排几个，这里要开三维，记录上次和上上次的状态，再一一判定，状态转移方程为 dp[i][j][k] = max(dp[i][j][k],dp[i - 1][k][t] + num[j])

代码：

#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define ll long long
const int N = 500+5;
const int MOD = 100000000;
const int INF = 0x3f3f3f3f;
using namespace std;
int n,m,top;
int cur[110],state[110],dp[110][110][110],num[110];
void init(){
    top = 0;
    int tot = 1 << m;
    for(int i = 0;i < tot;i++){
        if(i&(i<<1) || i&(i<<2)) continue;
        state[++top] = i;
        for(int j = 0;j < m;j++){
            if(i&1<<j) num[top]++;
        }
    }
}
int main(){
    char s[12];
    scanf("%d%d",&n,&m);
    memset(dp,0,sizeof(dp));
    memset(num,0,sizeof(num));
    memset(cur,0,sizeof(cur));
    memset(state,0,sizeof(state));
    init();
    for(int i = 1;i <= n;i++){
        scanf("%s",s + 1);
        cur[i] = 0;
        for(int j = 1;j <= m;j++){
            if(s[j] == 'H'){
                cur[i] |= 1<<(j - 1);
            }
        }
    }
    for(int i = 1;i <= top;i++){
        if(cur[1]&state[i]) continue;
        dp[1][i][1] = num[i];
    }
    for(int i = 1;i <= top;i++){
        if(cur[2]&state[i]) continue;
        for(int j = 1;j <= top;j++){
            if(cur[1]&state[j]) continue;
            if(state[i]&state[j]) continue;
            for(int k = 1;k <= top;k++){
                dp[2][i][j] = max(dp[2][i][j],dp[1][j][k] + num[i]);
            }
        }
    }
    for(int i = 3;i <= n;i++){
        for(int j = 1;j <= top;j++){
            if(cur[i]&state[j]) continue;
            for(int k = 1;k <= top;k++){
                if(cur[i - 1]&state[k]) continue;
                if(state[j]&state[k]) continue;
                for(int t = 1;t <= top;t++){
                    if(cur[i - 2]&state[t]) continue;
                    if(state[k]&state[t]) continue;
                    if(state[j]&state[t]) continue;
                    dp[i][j][k] = max(dp[i][j][k],dp[i - 1][k][t] + num[j]);
                }
            }
        }
    }
    ll ans = 0;
    for(int i = 1;i <= top;i++){
        for(int j = 1;j <= top;j++){
            if(dp[n][i][j] > ans) ans = dp[n][i][j];
        }
    }
    printf("%lld
",ans);
    return 0;
}