poj3254 状态压缩dp-Corn Fields

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 3
1 1 1
0 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.

题意：

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在n*m的土地上种玉米，1对应的土地可以种，0对应的土地不可以种，两个玉米地不能上下左右相邻，问有多少种种法。注意可以不种！

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题解：

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一道状态压缩dp的问题，可以先网上了解一下状态压缩dp。
每一行最多12个，那么最多用2的12次方就可以表示每一行的所有状态。
（1）对于每一行的状态j，满足j&(j-1) == 0
（2）对于相邻两行的状态j、k，满足（j&k)==0
（3）dp[i][j]表示第i行处于j状态时的方法数，状态转移式为：
dp[i][num[j]] += sum(dp[i-1][num[k]]);

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#include <iostream>
#include <string.h>
#include <cmath>
#define MAXN 4100
using namespace std;

int m,n;
int top;
int a[13];
int num[MAXN];
int dp[13][MAXN];

int isok(int x)
{
    if(x&(x<<1))
        return 0;
    return 1;
}

void cal(int n)
{
    top = 0;
    for(int i = 0; i < (1<<n); i++)
    {
        if(isok(i))
            num[top++] = i;
    }
}

int main()
{
    int b;
    while(cin >> m >> n)
    {
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        cal(n);
        for(int i = 1; i <= m; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                cin>>b;
                a[i] += b<<(n-j);
            }
        }
        for(int i = 1; i <= m; i++)
        {
            for(int j = 0; j < top; j++)
            {
                if(i == 1 && (num[j]|a[i])<=a[i])
                    dp[i][num[j]]++;
                else
                {
                    if((num[j]|a[i])<=a[i])
                    {
                        for(int k = 0; k < top; k++)
                        {
                            if(!(num[j]&num[k]) && (num[k]|a[i-1])<=a[i-1])
                                dp[i][num[j]] += dp[i-1][num[k]];
                        }
                    }
                }
            }
        }
        int ans = 0;
        for(int i = 0; i < top; i++)
        {
            ans += dp[m][num[i]];
        }
        cout << ans%100000000 << endl;
    }
    return 0;
}