POJ3254：Corn Fields（状压dp入门）

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Farmer John面对一块M*N的矩形牧场，希望在肥沃的土地上种植玉米供奶牛食用，但为了避免奶牛过于接近，他需要确保种植的地块不相邻。本问题探讨如何计算在给定地形条件下，符合要求的种植方案数量。

Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14098		Accepted: 7396

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

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

dp[i][j]表示第i行的第j种布局有多少种。dp[i][j] += dp[i-1][t]，t∈[0, 1<<m)，其中j布局要符合第i行地形，t布局要符合第i-1行的地形，j布局与t布局又合法（没有上下相邻的1）。

# include <stdio.h>
# include <string.h>
# define MOD 100000000
int map[13], val[1<<13], dp[13][1<<13];
bool judge(int i, int j) {return i&j;}
int main()
{
    int k, m, n, tmp, sum;
    while(~scanf("%d%d",&n,&m))
    {
        memset(map, 0, sizeof(map));
        memset(dp, 0, sizeof(dp));
        for(int i=1; i<=n; ++i)
            for(int j=1; j<=m; ++j)
            {
                scanf("%d",&tmp);
                if(!tmp)
                    map[i] |= 1<<(j-1);//模拟地形布局
            }
        int up = 1<<m;
        sum = k = 0;
        for(int i=0; i<up; ++i)
            if(!judge(i, i<<1))
                val[k++] = i;//存储所有可能的合法布局
        for(int i=0; i<k; ++i)
            if(!judge(map[1], val[i]))//由于map将贫瘠地存为1，此处判断该布局有无与贫瘠地重合
                dp[1][i] = 1; //无重合就记为1种方案
        for(int i=2; i<=n; ++i)
            for(int j=0; j<k; ++j)
                if(!judge(map[i], val[j]))
                    for(int t=0; t<k; ++t)
                        if(dp[i-1][t] && !judge(val[t], val[j]))
                            dp[i][j] += dp[i-1][t];
        for(int i=0; i<k; ++i)
            sum = (sum + dp[n][i]) % MOD;
        printf("%d\n",sum);
    }
    return 0;
}