POJ3254 Corn Fields(状压DP)

最新推荐文章于 2019-11-24 23:57:32 发布

原创最新推荐文章于 2019-11-24 23:57:32 发布 · 309 阅读

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本文介绍了一道关于在限定条件下计算农田种植方案数量的问题。通过动态规划和位运算的方法，解决了一个m*n网格中，如何在考虑土地肥沃性和避免相邻种植的限制下，计算出所有可能的种植方式。

Corn Fields

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16023		Accepted: 8447

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

题意：有一块 m * n 的农田，其中 1 能用来放牧，而 0 不能，且任意两块相邻的田不能一起放牧，问一共有多少种放牧方法（全不放牧也是一种方法）

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

const int N = 13;
const int Mod = 100000000;
int n, m;
int a[N], dp[N][1<<N];

bool judge1(int x){
    return !(x & (x>>1));
}

bool judge2(int x, int y){
    return !(x & y);
}

int main(){
    while(scanf("%d%d", &n, &m) == 2){
        memset(a, 0, sizeof(a));
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                int x;
                scanf("%d", &x);
                if(x) a[i] = a[i] | (1 << j);
            }
        }
        for(int i = 0; i < (1<<m); i++){
            if((a[0] | i) == a[0] && judge1(i))
                dp[0][i] = 1;
        }
        for(int i = 1; i < n; i++){
            for(int j = 0; j < (1<<m); j++){
                if((a[i] | j) == a[i] && judge1(j)){
                    for(int k = 0; k < (1<<m); k++){
                        if(judge2(j, k)){
                            dp[i][j] = (dp[i][j] + dp[i-1][k]) % Mod;
                        }
                    }
                }
            }
        }
        int ans = 0;
        for(int i = 0; i < (1<<m); i++){
            ans = (ans + dp[n-1][i]) %Mod;
        }
        printf("%d\n", ans);
    }
}

总结： 1. 注意位运算运算顺序，养成加（）的好习惯

2. 看到数据范围很小（尤其是小于32的，优先往二进制、状压这方面想）