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poj 1185 炮兵阵地

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题目描述

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

状压dp经典题

Solution

能够发现一行的状态远不止1024，所以可以把这东西扔到状态里即可

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#include <iostream>
#include <cstdio>
#define maxn 110
#define maxm 11 
using namespace std;

int n, m, g[maxn]; 

int st[100], c1, cnt[110];

int f[maxn][100][100];

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        char s[maxm]; scanf("%s", s + 1); 
        for (int j = 1; j <= m; ++j)
            if (s[j] == 'H') g[i] |= 1 << m - j; 
    }
    for (int i = 0; i < (1 << m); ++i) {
        if ((i << 1 & i) || (i << 2 & i)) continue;
        st[++c1] = i;
        for (int j = 1; j <= m; ++j) cnt[c1] += i >> j - 1 & 1; 
    }
    for (int i = 1; i <= c1; ++i) {
        if (g[1] & st[i]) continue;
        f[1][i][i] = cnt[i];
    }
    for (int i = 1; i < n; ++i)
        for (int j = 1; j <= c1; ++j) 
            for (int k = 1; k <= c1; ++k) {
                if (!f[i][j][k]) continue;
                for (int o = 1; o <= c1; ++o) {
                    if ((st[j] & st[o]) || (st[k] & st[o]) || (g[i + 1] & st[o])) continue;
                    f[i + 1][o][j] = max(f[i + 1][o][j], f[i][j][k] + cnt[o]); 
                }
            }
    int ans = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= c1; ++j)
            for (int k = 1; k <= c1; ++k) ans = max(ans, f[i][j][k]); 
    cout << ans << endl;
    return 0;
}