这是第一次写状态压缩。
题目描述：给你n*m的一个草地，放牛。牛不能相邻。并且值为0的地方不能放牛。问一共有多少种方案，可以放牛。
不放也算一种。
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 6143 | Accepted: 3259 |
Description
Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares
are infertile and can't be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice
as to which squares to plant.
Being a very open-minded man, Farmer John wants to consider all possible options for how to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways
he can choose the squares to plant.
Input
Lines 2..M+1: Line i+1 describes row i of the pasture with N space-separated integers indicating whether a square is fertile (1 for fertile, 0 for infertile)
Output
Sample Input
2 3 1 1 1 0 1 0
Sample Output
9
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 1000100
//#define LL __int64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
const int maxn = 10010;
using namespace std;
int mp[100][100];
int dp[15][1<<13];
LL Mod = 100000000;
int main()
{
int n, m;
while(cin >>n>>m)
{
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
cin >>mp[i][j];
for(int i = 0; i <= (1<<m); i++) dp[1][i] = 1;
for(int i = 1; i < n; i++)
{
for(int j = 0; j < (1<<m); j++)
{
int cnt = 0;
int k;
for(k = 1; k <= m; k++)//在行中找满足条件的
{
if(!mp[i][k] && (j&(1<<(k-1))))
break;
if(j&(1<<(k-1)))
cnt++;
else
cnt = 0;
if(cnt >= 2)
break;
}
if(k <= m) continue;
for(k = 0; k < (1<<m); k++)//在列中找满足条件的
{
int l;
for(l = 0; l < m; l++)
if((j&(1<<l)) && (k&(1<<l)))
break;
if(l < m) continue;
dp[i+1][k] += dp[i][j];
}
}
}
LL ans = 0;
for(int i = 0; i < (1<<m); i++)
{
int cnt = 0;
int k;
for(k = 1; k <= m; k++)
{
if(!mp[n][k] && (i&(1<<(k-1))))
break;
if(i&(1<<(k-1)))
cnt++;
else
cnt = 0;
if(cnt >= 2)
break;
}
if(k <= m) continue;
ans += dp[n][i];
ans %= Mod;
}
cout<<ans<<endl;
}
return 0;
}