POJ 3254 Corn Fields（状态压缩）

一道状态压缩的题,错了好多次....应该先把满足的情况预处理出来

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int fitnum,n,m;
int maps[],state[<<];
int dp[][<<];
#define mod 1000000000
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(maps,,sizeof(maps));
        fitnum = ;
        for(int i = ;i <= n;i++)
        {
            for(int j = ;j <= m;j++)
            {
                int s;
                scanf("%d",&s);
                if(!s)
                maps[i] += <<(m-j);
            }
        }
        memset(dp,,sizeof(dp));
        memset(state,,sizeof(state));
        for(int i = ;i < (<<m);i++)
        {
            if((i&(i<<)) == )
            state[fitnum++] = i;
        }
        for(int i = ;i < fitnum;i++)
        {
            if(!(state[i] & maps[]))
                dp[][i] = ;
        }
        for(int i = ;i <= n;i++)
        {
            for(int k = ;k < fitnum;k++)
            {
                if(state[k] & maps[i]) continue;
                for(int j = ;j < fitnum;j++)
                {
                    if(state[j] & maps[i-]) continue;
                    if(state[j] & state[k]) continue;
                    dp[i][k] = (dp[i][k]%mod + dp[i-][j]%mod)%mod;
                }
            }
        }
        int ans = ;
        for(int i = ;i < fitnum;i++)
        {
            ans = (ans%mod + dp[n][i]%mod) % mod;
        }
        printf("%d\n",ans);
    }
    return ;
}