首页 > 专题系列 > Java解POJ > POJ 3254 Corn Fields [解题报告] Java
2013
11-12

POJ 3254 Corn Fields [解题报告] Java

Corn Fields

问题描述 :

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.

输入:

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)

输出:

Line 1: One integer: the number of ways that FJ can choose the squares modulo 100,000,000.

样例输入:

2 3
1 1 1
0 1 0

样例输出:

9

温馨提示:

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.

解题代码:

import java.util.ArrayList;  
    import java.util.List;  
    import java.util.Scanner;  
      
    public class Main{  
      
        static int n,m;  
        static List< Integer> num[] = new ArrayList[13];  
        static int dp[][] = new int[13][1024];  
          
        public static void main(String[] args) {  
              
            Scanner scan = new Scanner(System.in);  
              
            m = scan.nextInt();  
            n = scan.nextInt();  
              
            for(int i=0;i< 13;i++)  
                num[i] = new ArrayList< Integer>();  
              
            for(int i=0;i< m;i++){  
                int temp = 0;  
                for(int j=0;j< n;j++){      
                    int a = scan.nextInt();  
                    a = 1 - a;  
                    temp = temp*2+a;  
                }  
                getNum(i,temp);  
            }  
              
            for(int i=0;i< num[0].size();i++)  
                dp[0][i] = 1;  
              
            for(int i=1;i< m;i++){  
                for(int j=0;j< num[i].size();j++){  
                    for(int k=0;k< num[i-1].size();k++){  
                        if((num[i].get(j)&num[i-1].get(k))!=0)  
                            continue;  
                        dp[i][j] += dp[i-1][k];  
                    }  
                }  
            }  
              
            int ans = 0;  
            for(int j=0;j< num[m-1].size();j++){  
                ans += dp[m-1][j];  
                ans %= 1000000000;  
            }  
              
            System.out.println(ans);  
      
        }  
      
        public static void getNum(int r, int temp) {  
              
            for(int i=0;i<(1<< n);i++){  
                if(((i<< 1)&i)!=0||((i>>1)&i)!=0) //处理连续的1  
                    continue;  
                if((i&temp)!=0)  
                    continue;  // 处理 原来 为0的位置   不能放牛  
                num[r].add(i);  
            }  
              
        }  
      
    }

  1. #include <stdio.h>
    int main(void)
    {
    int arr[] = {10,20,30,40,50,60};
    int *p=arr;
    printf("%d,%d,",*p++,*++p);
    printf("%d,%d,%d",*p,*p++,*++p);
    return 0;
    }

    为什么是 20,20,50,40,50. 我觉得的应该是 20,20,40,40,50 . 谁能解释下?