<h1 id="heading">题目</h1>
<blockquote>
<blockquote>
We have a grid of size N × 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 × 4: 
<div class='image'><a href='//static.oyohyee.com/post/uva11464.png' target='_blank' rel='noopener noreferrer'><img src="//static.oyohyee.com/post/uva11464.png" /></a></div> 
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.
 
</blockquote>
<h2 id="input">Input</h2>
<blockquote>
The first line of input is an integer T (T &lt; 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.
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
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.
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
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
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
Case 1: 0 
Case 2: 3 
Case 3: -1
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
本来想使用动态规划,仔细一想压根没法计算
枚举 第一行的所有可能 
模拟 生成剩下的每一行,计算与原图最 相近 的
输出结果即可,没什么明显的大坑,就是交题的时候服务器抽了2个多小时
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;
#include &lt;iomanip&gt;
#include &lt;iostream&gt;
#include &lt;map&gt;
#include &lt;set&gt;
#include &lt;list&gt;
#include &lt;queue&gt;
#include &lt;stack&gt;
#include &lt;string&gt;
#include &lt;vector&gt;
#include &lt;bitset&gt;
#include &lt;functional&gt;

using namespace std;

const int INF = 0x7FFFFFFF;
const double eps = 1e-10;

const int maxn = 16;
const int delta[3][2] = {{-1,-1},{-1,1},{-2,0}};


int state[maxn];
int Map[maxn];

int n;
int kase = 1;

//比较两种状态,返回需要变成1的0的数量,不能完成返回-1
int Get(int res,int e) {
 int ans = 0;
 for(int i = 0;i &lt; n;i++) {
 int a = res &amp; 1;
 int b = e &amp; 1;

 res &gt;&gt;= 1;
 e &gt;&gt;= 1;

 if(a != b)
 if(a == 0)
 ans++;
 else
 return -1;
 }
 return ans;
}

inline int SetI(int num,int i,bool flag) {
 if(flag)
 return num | (1 &lt;&lt; i);
 else
 return num &amp; (~(1 &lt;&lt; i));
}

inline bool GetI(int num,int i) {
 return num &gt;&gt; i &amp; 1;
}

void Do() {
 cin &gt;&gt; n;
 int Min = INF;

 for(int i = 0;i &lt; n;i++) {
 int m = 0;
 for(int j = 0;j &lt; n;j++) {
 int t;
 cin &gt;&gt; t;
 m &lt;&lt;= 1;
 m = SetI(m,0,t);
 }
 Map[i] = m;
 }

 for(int k = 0;k &lt; (1 &lt;&lt; n);k++) {
 //枚举初始态
 int ans = 0;
 state[0] = k;

 int temp = Get(Map[0],state[0]);

 if(temp == -1)
 continue;
 else
 ans += temp;

 for(int i = 1;i &lt; n;i++) {
 //处理第i行
 for(int j = 0;j &lt; n;j++) {
 //处理第j个位置
 int sum = 0;
 for(int t = 0;t &lt; 3;t++) {
 int xx = i + delta[t][0];
 int yy = j + delta[t][1];
 if(xx &gt;= 0 &amp;&amp; yy &gt;= 0 &amp;&amp; xx &lt; n &amp;&amp; yy &lt; n) {
 sum += GetI(state[xx],yy);
 }
 }
 if(sum &amp; 1)
 state[i] = SetI(state[i],j,1);
 else
 state[i] = SetI(state[i],j,0);
 }

 temp = Get(Map[i],state[i]);
 if(temp == -1)
 break;
 else
 ans += temp;
 }

 if(temp == -1)
 continue;
 else 
 Min = min(Min,ans);
 }

 cout &lt;&lt; &#34;Case &#34; &lt;&lt; kase++ &lt;&lt; &#34;: &#34;;
 if(Min == INF)
 cout &lt;&lt; -1 &lt;&lt; endl;
 else
 cout &lt;&lt; Min &lt;&lt; endl;
}

int main() {
 cin.tie(0);
 cin.sync_with_stdio(false);

 int T;
 cin &gt;&gt; T;
 while(T--)
 Do();

 return 0;
}
</pre>

Uva 11464.Even Parity