本文主要是介绍[ACM] POJ 3254 Corn Fields(状态压缩），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 8062 | Accepted: 4295 |
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
Hint
1 2 34
There are four ways to plant only on one squares (1, 2, 3, or 4), three ways to plant on two squares (13, 14, or 34), 1 way to plant on three squares (134), and one way to plant on no squares. 4+3+1+1=9.
Source
解题思路：
题意为有n行m列的长方形，分成n*m个格子，每个格子标记为0或1，在这些格子里面放牛，其中1代表该格子可以放牛，0代表不能放牛，且相邻的两个格子不能同时有牛，问总的方案数。
思想为状态压缩，把每一行放牛的状态看作一个二进制数，dp[i][j]表示第i行状态为j时前i行共有的方案数。
代码：
#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
const int mod=100000000;
const int maxn=12;
int dp[maxn+1][(1<<maxn)+1];
int num[maxn+1];
int n,m;bool check(int i,int x)//检查第i行出现状态x是否合法
{if((x&num[i])!=x)//很巧妙,判断第i行出现的状态x是否合法，为什么这么写，因为0的地方不能放牛，//合法状态与原始状态0位且为0，原始状态1位的地方与合法状态对应位且都等于合法状态位上的数字return 0;if(x&(x>>1)||x&(x<<1))//不能有相邻的两个1return 0;return 1;
}int main()
{cin>>n>>m;int x;memset(num,0,sizeof(num));memset(dp,0,sizeof(dp));for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){cin>>x;if(x)num[i]=num[i]|(1<<(j-1));//把每一行的状态保存到num[i]中去}int Max=(1<<m);dp[0][0]=1;//注意这一句啊！for(int i=1;i<=n;i++)//枚举每一行{for(int j=0;j<Max;j++){if(!check(i,j))continue;for(int k=0;k<Max;k++)if((j&k)==0){dp[i][j]+=dp[i-1][k];if(dp[i][j]>=mod)dp[i][j]%=mod;}}}int ans=0;for(int i=0;i<Max;i++){ans+=dp[n][i];if(ans>=mod)ans%=mod;}cout<<ans;return 0;
}
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