HDU1693 Eat the Trees 插头DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1693

　　插头DP的入门题，属于轮廓动态规划一类，推荐看《基于连通性状态压缩的动态规划问题》，陈丹琦写的，Orz女神...

//STATUS:C++_AC_0MS_272KB
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
using namespace std;
#define LL __int64
#define pii pair<int,int>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int N=12,M=100000,INF=0x3f3f3f3f,MOD=100000000;
const double DNF=100000000000;

int ma[N][N];
LL f[2][1<<N];
int T,n,m;

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,k,x,y,up,icase=0,p;
    scanf("%d",&T);
    while(T--)
    {
        mem(f,0);
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
                scanf("%d",&ma[i][j]);
        f[0][0]=p=1;
        up=1<<(m+1);
        for(i=0;i<n;i++,mem(f[p=!p],0)){
            for(j=0;j<m;j++,mem(f[p=!p],0)){
                x=1<<j;
                y=1<<(j+1);
                for(k=0;k<up;k++){
                    if(ma[i][j]){
                        f[p][k^x^y]+=f[!p][k];
                        if((k&x) && (k&y))continue;
                        if((k&x)^(k&y))f[p][k]+=f[!p][k];
                    }
                    else if(!(k&x) && !(k&y)){
                        f[p][k]+=f[!p][k];
                    }
                }
            }
            for(j=0;j<(1<<m);j++)f[p][j<<1]=f[!p][j];
        }

        printf("Case %d: There are %I64d ways to eat the trees.\n",++icase,f[!p][0]);
    }
    return 0;
}