Even Parity UVA 11464

http://acm.uva.es/local/online_judge/search_uva.html

这一题需要遍历所有的可能性，从第一行开始有2^n种可能，接下来的n-1行可以由第一行推测出来。

主要是如何遍历的问题，我想了一个下午都没得到正确的答案，索性全部删除，借用书上的思路把代码敲出来，为什么一模一样呢？因为已经到了该无可改的地步了。

精彩之处1.使用移位运算枚举第一行的所有可能性。2.比较a矩阵和b矩阵计算出要改变的零的个数。3.计算周围元素之和的sum条件语句也是非常精炼的。

/*
*1.枚举第一行的每个元素
*2.计算推测第二行至第n行的元素
*3.计算矩阵的不同元素个数获取ct值
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#define INF 100000000
using namespace std;
int a[20][20];
int b[20][20];
int n;
int check(int s)
{
    int ct=0;
    memset(b,0,sizeof(b));
    for(int i=0;i<n;i++)
    {
       if(s&(1<<i))b[0][i]=1;
       else if(a[0][i]==1)return INF;
    }
    for(int i=1;i<n;i++)//从第二行开始　
    {
        for(int j=0;j<n;j++)
        {
            int sum=0;
            if(i>1)sum+=b[i-2][j];
            if(j>0)sum+=b[i-1][j-1];
            if(j<n-1)sum+=b[i-1][j+1];
            b[i][j]=sum%2;
            if(a[i][j]==1&&b[i][j]==0)return INF;
        }
    }
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<n;j++)
        {
            if(a[i][j]!=b[i][j])ct++;
        }
    }
    return ct;
}
int main()
{
   int T;
   cin>>T;
   for(int cas=0;cas<T;cas++)
   {
      cin>>n;
      for(int i=0;i<n;i++)
      {
        for(int j=0;j<n;j++)
        {
            cin>>a[i][j];
        }
      }
      int minct=INF;
      for(int s=0;s<(1<<n);s++)
      {
        int t=check(s);
        if(t<minct)minct=t;
      }
      if(minct==INF)
      {
        printf("Case %d: %d\n",cas+1,-1);
      }
      else printf("Case %d: %d\n",cas+1,minct);
   }
}

题目在这里：

D

Even  Parity Input: Standard Input Output: Standard Output

 

We have a grid of size N x N. Each cell of the grid initially contains a zero(0) or a one(1).
The parity of a cell is the number of 1s surrounding that cell. A cell is surrounded by at most 4 cells (top, bottom, left, right).

Suppose we have a grid of size 4 x 4: 

 

 

1	0	1	0	The parity of each cell would be	1	3	1	2
1	1	1	1		2	3	3	1
0	1	0	0		2	1	2	1
0	0	0	0		0	1	0	0

 

 

 

 

 

 

 

For this problem, you have to change some of the 0s to 1s so that the parity of every cell becomes even. We are interested in the minimum number of transformations of 0 to 1 that is needed to achieve the desired requirement.

 

Input

The first line of input is an integer T (T<30) that indicates the number of test cases. Each case starts with a positive integer N(1≤N≤15). Each of the next N lines contain N integers (0/1) each. The integers are separated by a single space character.

 

Output

For each case, output the case number followed by the minimum number of transformations required. If it's impossible to achieve the desired result, then output -1 instead.

 

Sample Input                             Output for Sample Input

 

3 3 0 0 0 0 0 0 0 0 0 3 0 0 0 1 0 0 0 0 0 3 1 1 1 1 1 1 0 0 0

Case 1: 0 Case 2: 3 Case 3: -1

 

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Problem Setter: Sohel Hafiz,

Special Thanks: Derek Kisman, Md. Arifuzzaman Arif