状压DP POJ 3254 Corn Fields

题目传送门

/*
    状态压缩DP：先处理硬性条件即不能种植的，然后处理左右不相邻的，
    接着就是相邻两行查询所有可行的种数并累加

    写错一个地方差错N久:)
    详细解释：http://www.tuicool.com/articles/JVzMVj
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

const int MAXN = 15;
const int INF = 0x3f3f3f3f;
const int MOD = 1e8;
int dp[MAXN][1 << MAXN];
int st[1 << MAXN];
int rem[1 << MAXN];

int solve(int m)
{
    int cnt = 0;
    int tot = 1 << m;
    for (int i=0; i<tot; ++i)
    {
        if ((i & (i << 1)) == 0)
        {
            st[++cnt] = i;
        }
    }

    return cnt;
}

void work(int n, int m, int cnt)
{
    memset (dp, 0, sizeof (dp));
    for (int i=1; i<=cnt; ++i)
    {
        if ((st[i] & rem[1]) == 0)
        {
            dp[1][i] = 1;
        }
    }

    for (int i=2; i<=n; ++i)
    {
        for (int k=1; k<=cnt; ++k)
        {
            if ((st[k] & rem[i]) == 0)
            {
                for (int j=1; j<=cnt; ++j)
                {
                    if (((st[j] & rem[i-1]) == 0) && ((st[j] & st[k]) == 0))
                        dp[i][k] = (dp[i][k] + dp[i-1][j]) % MOD;
                }
            }
        }
    }

    int ans = 0;
    for (int i=1; i<=cnt; ++i)
    {
        ans = (ans + dp[n][i]) % MOD;
    }

    printf ("%d
", ans);
}

int main(void)        //POJ 3254 Corn Fields
{
    #ifndef ONLINE_JUDGE
    freopen ("F.in", "r", stdin);
    #endif
    
    int n, m;
    while (~scanf ("%d%d", &n, &m))
    {
        int cnt = solve (m);

        memset (rem, 0, sizeof (rem));
        for (int i=1; i<=n; ++i)
        {
            int x;
            for (int j=1; j<=m; ++j)        //看错n和m
            {
                scanf ("%d", &x);
                if (x == 0)
                {
                    rem[i] = rem[i] | (1 << (m-j));
                }
            }
        }

        work (n, m, cnt);
    }

    return 0;
}