B. Seating of Students
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Students went into a class to write a test and sat in some way. The teacher thought: "Probably they sat in this order to copy works of each other. I need to rearrange them in such a way that students that were neighbors are not neighbors in a new seating."

The class can be represented as a matrix with n rows and m columns with a student in each cell. Two students are neighbors if cells in which they sit have a common side.

Let's enumerate students from 1 to n·m in order of rows. So a student who initially sits in the cell in row i and column j has a number(i - 1)·m + j. You have to find a matrix with n rows and m columns in which all numbers from 1 to n·m appear exactly once and adjacent numbers in the original matrix are not adjacent in it, or determine that there is no such matrix.

Input

The only line contains two integers n and m (1 ≤ n, m ≤ 105; n·m ≤ 105) — the number of rows and the number of columns in the required matrix.

Output

If there is no such matrix, output "NO" (without quotes).

Otherwise in the first line output "YES" (without quotes), and in the next n lines output m integers which form the required matrix.

Examples
input
Copy 
2 4
output
Copy 
YES
5 4 7 2 
3 6 1 8
input
Copy 
2 1
output
Copy 
NO
Note

In the first test case the matrix initially looks like this:

1 2 3 4
5 6 7 8

It's easy to see that there are no two students that are adjacent in both matrices.

In the second test case there are only two possible seatings and in both of them students with numbers 1 and 2 are neighbors.

            这道题应该是用一种dfs来做搜索比较好~详细情况代码中解释。

#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
#include<vector>
using namespace std;
vector<int>a, b;
int k[100005];//初始情况~~最后将是答案
int fx[4][2] = { 1,0,0,1,-1,0,0,-1 };
int n, m;
bool check(int a, int b)
{
	int x = (a - 1) / m + 1, y = (a - 1) % m + 1;
	for (int s = 0; s < 4; s++)
	{
		int x1 = x + fx[s][0];
		int y1 = y + fx[s][1];
		if ((x1 - 1)*m + y1 == k[b])
		{
			return 1;
		}
	}
	return 0;
}
bool dp(int x)//代表x及x位之后能否匹配成功
{
	if (x >= n*m + 1)//超过n*m则表示之前的全部匹配成功了
	{
		return 1;
	}
	int xx = (x - 1) / m + 1;
	int yy = (x - 1) % m + 1;
	for (int e = x ; e <= n*m; e++)
	{
		swap(k[x], k[e]);
		if ((xx != 1 && check(k[x], (xx - 2)*m + yy)) || (yy != 1 && check(k[x], (xx - 1)*m + yy - 1)))
		{//分别检查上方及左方的数是否和最初的时候临近
			continue;
		}
		if (dp(x + 1))
		{//如果成功的话进行下次搜索
		//	cout << x << endl;
			return 1;
		}
		swap(k[e], k[x]);//恢复原来情况；
	}
	return 0;
}
int main()
{
	scanf("%d%d", &n, &m);
	for (int s = 1; s <= n*m; s++)
	{
		k[s] = s;
	}
	if (dp(1))
	{
		printf("YES\n");
		for (int s = 1; s <= n; s++)
		{
			printf("%d", k[(s - 1)*m + 1]);
			for (int e = 2; e <= m; e++)
			{
				printf(" %d", k[(s - 1)*m + e]);
			}
			printf("\n");
		}
	}
	else
	{
		printf("NO\n");
	}
	return 0;
}