poj3254

题目大意：给你一个n*m的矩形，里面有些格子可以选取。。在可选取选取格子，要求不能相邻。。求方案数。。

思路:状态压缩dp

       f[i][j]表示第i行状态为J的方案数

      f[i][j] = sigma(f[i - 1][k])  状态k表示的取法与j取的不相邻。且j为合法取法（与给定的不冲突）

   

/*
  Time:2013-03-19 23:07:07
  State:Accepted
*/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <algorithm>
using namespace std;
int n, m , map[25] , tot(0), a[1000] , maxnum;
int  f[15][1 << 14];
void init(){
     int x;
     scanf("%d%d", &n ,&m);
     for (int i = 1;  i <= n; ++i)
         for (int j = 0; j < m; ++j){
              scanf("%d",&x);
              map[i] = map[i] | (x << j);
         }
}

void work_opt(int i , int opt){
     if (i > m){
         a[++tot] = opt;  
         return;
     }
     work_opt(i + 1, opt);
     work_opt(i + 2, opt | (1 << i - 1));
}

void dp(){
     int opt;
     work_opt(1,0);
     f[0][0] = 1;
     for (int i = 1; i <= n; ++i)
         for (int j = 1; j <= tot; ++j){
            opt = map[i] & a[j];
            if (opt != a[j]) continue;
            for (int k = 1; k <= tot; ++k)
               if ((opt & a[k]) == 0) f[i][opt] = (f[i][opt] + f[i-1][a[k]])% 100000000;      
         } 
     int  ans = 0;
    /* for (int i = 1; i <= tot; ++i)
         printf("a[%d] = %d\n",i , a[i]);
     for (int i = 1; i <= n; ++i)
         for (int  j = 1 ; j <= tot; ++j)
              printf("f[%d][%d] = %d\n", i, a[j], f[i][a[j]]);*/
     for (int i = 1; i <= tot; ++i)
         ans = (ans + f[n][a[i]]) % 100000000;
     printf("%d\n", ans);
}

int main(){
    freopen("poj3254.in","r",stdin);    
    freopen("poj3254.out","w",stdout);
    init();
    dp();
    fclose(stdin); fclose(stdout);
}