[POJ]3254 Corn Fields [状压dp]

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本文介绍了一个经典的编程竞赛问题——CornFields。Farmer John要在一块M*N的矩形牧场中种植玉米，但由于土地肥沃度不一及考虑牛吃草的习惯，不能选择相邻的土地种植。文章提供了详细的解题思路及代码实现，并解释了如何使用动态规划来解决这个问题。

Corn Fields
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 16318 Accepted: 8615
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.
Source

USACO 2006 November Gold

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int MOD = (int)1e9;
int dp[12][1<<10], ste[1<<10], sz, fid[12], n, m;

void slove() {
    sz = 0;
    for(int i = 0; i < 1<<m; ++i) {
        if(i&i<<1) continue;
        if(!(i&fid[0])) dp[0][sz] = 1;
        ste[sz++] = i;
    }
    for(int i = 1; i < n; ++i) {
        for(int j = 0; j < sz; ++j) {
            if(ste[j]&fid[i]) continue;
            for(int k = 0; k < sz; ++k) {
                if(ste[j]&ste[k]) continue;
                dp[i][j] += dp[i - 1][k];
                dp[i][j] %= MOD;
            }
        }
    }
    int ans = 0;
    for(int i = 0; i < sz; ++i) {
        ans += dp[n - 1][i];
        ans %= MOD;
    }
    printf("%d\n", ans);
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < m; ++j) {
            int x;
            scanf("%d", &x);
            if(!x) fid[i] += 1<<j;
        }
    }
    slove();
    return 0;
}