POJ 3254 Corn Fields | 状压DP

Posted 2020-10-20 MSPqwq

题目:

http://poj.org/problem?id=3254

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题解:

把一行压成一个状态

这样枚举每行,枚举这行和上行的状态,判断合法然后转移

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define N 15
typedef long long ll;
using namespace std;
ll m,n,f[N][1<<N],a[N][N],poss[N],t,ans;
inline bool ok(ll a,ll b)
{
    for (int i=N;i>=0;i--)
    if (!(a&(1<<i)) && (b&(1<<i))!=0 ) return 0;
    if (b & (b<<1)) return 0;
    return 1;
}
int main()
{
    scanf("%lld%lld",&n,&m);t=1<<m;
    for (int i=1;i<=n;i++)
    for (int j=0;j<m;j++)
    {
        scanf("%lld",&a[i][j]);
        if (a[i][j]==1) poss[i]+=(1<<j);
    }
    f[0][0]=1;
      for (int i=1;i<=n;i++)
    for (int j=0;j<t;j++)
        if (ok(poss[i],j))
        for (int k=0;k<t;k++)
            if ((j&k)==0) f[i][j]+=f[i-1][k];
    for (int i=0;i<t;i++)
    ans+=f[n][i];
    printf("%lld\n",ans%100000000);
    return 0;
}