poj_3254 Corn Fields（状压dp）

Corn Fields

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 14091 Accepted: 7393

Description

Farmer John has purchased a lush new rectangular pasture composed of M byN (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 andN
Lines 2..M+1: Line i+1 describes row i of the pasture withN 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

Hint

Number the squares as follows:

1 2 3  4  

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.

题目要求有多少种种树方案，满足

（1）数值为0的方格不能种树，

（2）相邻方格不能都种树。

用若干个整数ma[i]分别表示每一行的状态，二进制中1表示可以种树，0表示不可以种树。

同样用整数表示状态，设dp[i][j]为第i行选择状态为j时的所有方案数。

状态方程 dp[i][j] = sum(dp[i-1][k]) （（1）k满足不与j有都种树的相邻方格，

                                                               （2）j满足j中没有都种树的相邻方格，k也一样，

                                                               （3）j和k满足不在数值为0的方格上种树）

上面的状态方程中的

（1）可用 j&k 判断，为0即没有有相邻方格；

（2）可用 j&(j<<1) 判断，为0即没有相邻方格；

（3）可用 j&ma[i] 判断，为j即在没有在数值为0的方格上种树。

要注意不种树也算一种方案；记得取模。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define FOP2 freopen("data1.txt","w",stdout)#define inf 0x3f3f3f3f#define maxn 1000010#define mod 100000000#define PI acos(-1.0)#define LL long longusing namespace std;int n, m;int dp[20][1<<12];int ma[20];int main(){    //FOP;    while(~scanf("%d%d", &n, &m))    {        memset(ma, 0, sizeof(ma));        memset(dp, 0, sizeof(dp));        int f;        for(int i = 1; i <= n; i++)        {            for(int j = 0; j < m; j++)            {                scanf("%d", &f);                if(f) ma[i] |= 1<<j;            }        }        for(int j = 0; j < 1<<m; j++)        {            if((j&(j<<1)) || ((j&ma[1]) != j && j)) continue;            dp[1][j] = 1;        }        for(int i = 2; i <= n; i++)        {            for(int j = 0; j < 1<<m; j++)            {                if((j&(j<<1)) || ((j&ma[i]) != j && j)) continue;                for(int k = 0; k < 1<<m; k++)                {                    if(k&(k<<1) || (j&k)) continue;                    dp[i][j] = dp[i][j] + dp[i-1][k];                }            }        }        int ans = 0;        for(int i = 0; i < 1<<m; i++)  ans = (ans + dp[n][i])%mod;        printf("%d\n", ans);    }    return 0;}