题目链接

题目

Artem is building a new robot. He has a matrix $a$ consisting of $n$ rows and $m$ columns. The cell located on the $i$ -th row from the top and the $j$ -th column from the left has a value $a_{i,j}$ written in it.

If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side.

Artem wants to increment the values in some cells by one to make a good.

More formally, find a good matrix $b$ that satisfies the following condition —

For all valid $(i,j)$ , either $b_{i,j}=a_{i,j}$ or $b_{i,j}=a_{i,j}+1$ .

For the constraints of this problem, it can be shown that such a matrix $b$ always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.

题目大意

给定一个 $n$ 行 $m$ 列的矩阵 $a$ 。自顶向下第 $i$ 行和自左向右第 $j$ 列的单元格里有值 $a_{i，j}$ 。

如果没有两个相邻单元格包含相同的值，则矩阵是好的，如果两个单元格共享一条边，则称其为相邻。对于给定的矩阵 $a$ ，找到满足以下条件的矩阵 $b$ ：

矩阵 $b$ 是好的矩阵。
对于所有单元格 $(i,j)$ ，要么 $b_{i,j}=a_{i,j}$ ，要么 $b_{i,j}=a_{i,j}+1$ .

可以证明这样的矩阵 $b$ 总是存在的。如果有多个可能的答案，输出其中的任何一个，不必最小化增量的数量。

思路

我们的目标是让矩阵任意两个相邻的格子上的值不一样。容易发现我们的操作是独立为每个单元格选择是否增加 $1$ ，那么我们就可以独立的为每个单元格选择奇偶性。

我们可以将整个棋盘如下图所示进行黑白染色：

其中黑色格子放奇数，白色格子放偶数，就可以保证任意两个相邻的格子里的数字均不相同了。

代码

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
#include <math.h>
#include <map>
#include <queue>
#include <vector>
#include <stdlib.h>
#include <time.h>
using namespace std;
using LL=long long;
const int N=1e6+5;
//const LL mod=998244353;
const LL mod=1e9+7;
int n,m,k;
LL solve()
{
	scanf("%d%d",&n,&m);
	for (int i=1,x;i<=n;++i,printf("\n"))
		for (int j=1;j<=m;++j)
		{
			scanf("%d",&x);
			if ((i+j)&1)
				if (x&1) printf("%d ",x+1);
				else printf("%d ",x);
			else
				if (x&1) printf("%d ",x);
				else printf("%d ",x+1);
				
		}
	return 0;
}
int main()
{
	int T=1;
	scanf("%d",&T); 

	while (T--) solve();
	return 0;
}