[poj 3254] Corn Fields 状压dp

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：

　　农场主John有块M*N的田地，想为他的牛种一些谷物喂养他的小乳牛，但是有些地是荒废的，不能用来种植农作物，0表示荒废的，然而他还不想让他的牛（种植的农作物）在相邻的田地里，问总共有多少种种植的方案。

思路：

　　由于N不大，可以用状态压缩来表示每一行的种植状态，dp[i][j]表示种植到第i行为j状态时的方案数，只有i-1行有关。可得状态转移方程：

dp[i][j] =  Σdp[i-1][k]  k为i-1行所有满足i行状态j的状态

代码：

　　nowok()判断当行所有满足条件的状态，不能种在非种植区，也不能相邻。

　　preok()判断不能与上一行相邻。

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define mod 100000000
int M, N;
const int MaxN = (1<<12)+5;
const int MaxM = 12;
LL dp[MaxM][MaxN];
int farm[13][13];

bool GetBit(int i, int k) 
{
    return (1 & (i >> k));
}

bool nowok(int n, int r)
{
    for (int i = 0; i < N; i++) {
        if (farm[r][N-1-i] == 0 && GetBit(n, i) == 1)
            return 0;
    }
    for (int i = 0; i < 11; i++) {
        if (GetBit(n, i) == 1 && GetBit(n, i+1) == 1)
            return 0;
    }
    return 1;
}

bool preok(int p, int n)
{
    for (int i = 0; i < 12; i++) {
        if (GetBit(p, i) && GetBit(n, i))
            return 0;
    }
    return 1;
}

int main()
{
    //freopen("1.txt", "r", stdin);
    scanf("%d%d", &M, &N);
    for (int i = 0; i < M; i++)
        for (int j = 0; j < N; j++) {
            scanf("%d", &farm[i][j]);
        }

    int sp = 1<<N;
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < sp; i++) {
        if (nowok(i, 0))
            dp[0][i] = 1;
    }
    
    for (int i = 1; i < M; i++) {
        for (int j = 0; j < sp; j++) {
            if (nowok(j, i)) {
                for (int k = 0; k < sp; k++) {
                    if (dp[i-1][k] && preok(k, j)) {
                        dp[i][j] = (dp[i][j]+dp[i-1][k])%mod;
                        dp[i][j] %= mod;
                    }
                }
            }
        }
    }
    int ans = 0;
    for (int i = 0; i < sp; i++) {
        if (dp[M-1][i]) {
            ans = (ans+dp[M-1][i])%mod;
            ans %= mod;
        }
    }
    printf("%d
", ans);

    return 0;
}