POJ 3254 （基础 状态压缩 DP ）

Corn Fields

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 14837 Accepted: 7768

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

Line 1: Two space-separated integers: M and N 
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

Line 1: One integer: the number of ways that FJ can choose the squares modulo 100,000,000.

Sample Input

2 31 1 10 1 0

Sample Output

9

【题目大意】一个矩阵里有很多格子，每个格子有两种状态，可以放牧和不可以放牧，可以放牧用1表示，否则用0表示，在这块牧场放牛，要求两个相邻的方格不能同时放牛，即牛与牛不能相邻。问有多少种放牛方案(一头牛都不放也是一种方案)

分析：状态压缩DP，时间复杂度O（n* 2^m *2^m） , 用 x & （x<<1）来判断相邻位置是否有都为1的情况，

【状态表示】dp[state][i]：在状态为state时，到第i行符合条件的可以放牛的方案数

【状态转移方程】dp[state][i] =Sigma dp[state'][i-1] （state'为符合条件的所有状态）

【DP边界条件】首行放牛的方案数dp[state][1] =1(state符合条件) OR 0 (state不符合条件)

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <queue>#define mem(p,k) memset(p,k,sizeof(p));#define rep(a,b,c) for(int a=b;a<c;a++)#define pb push_back#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define inf 0x6fffffff#define ll long longusing namespace std;const int N=30010;const int mod=100000000;int n,m,num,tol;int mapp[15],plan[10000];int dp[15][10000];int main(){    while(~scanf("%d%d",&n,&m)){        tol=0;        for(int i=0;i<(1<<m);i++){            if(i&(i<<1))continue;            plan[tol++]=i;        }        for(int i=1;i<=n;i++){            mapp[i]=0;            for(int j=0;j<m;j++){                scanf("%d",&num);                if(!num)mapp[i]+=(1<<j);            }        }        for(int i=0;i<tol;i++){            if(plan[i] & mapp[1])dp[1][i]=0;            else dp[1][i]=1;        }        for(int i=2;i<=n;i++){            for(int j=0;j<tol;j++){                if(plan[j] & mapp[i])continue;                for(int k=0;k<tol;k++){                    if(plan[j] & plan[k] || plan[k] & mapp[i-1])continue;                    dp[i][j]+=dp[i-1][k];                    dp[i][j]%=mod;                }            }        }        int ans=0;        for(int i=0;i<tol;i++)ans=(ans+dp[n][i])%mod;        printf("%d\n",ans);    }    return 0;}/**/