Codeforces Round #385 (Div. 2)B. Hongcow Solves A Puzzle【思维+暴力】

B. Hongcow Solves A Puzzle

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Hongcow likes solving puzzles.

One day, Hongcow finds two identical puzzle pieces, with the instructions "make a rectangle" next to them. The pieces can be described by an nby m grid of characters, where the character 'X' denotes a part of the puzzle and '.' denotes an empty part of the grid. It is guaranteed that the puzzle pieces are one 4-connected piece. See the input format and samples for the exact details on how a jigsaw piece will be specified.

The puzzle pieces are very heavy, so Hongcow cannot rotate or flip the puzzle pieces. However, he is allowed to move them in any directions. The puzzle pieces also cannot overlap.

You are given as input the description of one of the pieces. Determine if it is possible to make a rectangle from two identical copies of the given input. The rectangle should be solid, i.e. there should be no empty holes inside it or on its border. Keep in mind that Hongcow is not allowed to flip or rotate pieces and they cannot overlap, i.e. no two 'X' from different pieces can share the same position.

Input

The first line of input will contain two integers n and m (1 ≤ n, m ≤ 500), the dimensions of the puzzle piece.

The next n lines will describe the jigsaw piece. Each line will have length m and will consist of characters '.' and 'X' only. 'X' corresponds to a part of the puzzle piece, '.' is an empty space.

It is guaranteed there is at least one 'X' character in the input and that the 'X' characters form a 4-connected region.

Output

Output "YES" if it is possible for Hongcow to make a rectangle. Output "NO" otherwise.

Examples

input

2 3
XXX
XXX

output

YES

input

2 2
.X
XX

output

NO

input

5 5
.....
..X..
.....
.....
.....

output

YES

Note

For the first sample, one example of a rectangle we can form is as follows

111222
111222

For the second sample, it is impossible to put two of those pieces without rotating or flipping to form a rectangle.

In the third sample, we can shift the first tile by one to the right, and then compose the following rectangle:

.....
..XX.
.....
.....
.....

题目大意：

给你一个图形，因为图形比较重，不能对其进行翻转，只能平移（两个图形不能重叠），问是否能够构造出来一个矩形。

思路：

看到什么四面性，感觉好复杂。其实我们因为不能对其进行翻转（旋转），那么我们必须原来的图形是矩形，才能使得两个图形拼在一起是一个矩形。

那么我们只要判断初始的图形是不是矩形即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
char a[505][505];
int vis[505][505];
int lie[505];
int n,m;
int Slove(int x,int y)
{
    int sum=-1;
    for(int i=x;i<n;i++)
    {
        if(a[i][y]=='X')
        {
            int cnt=0;
            for(int j=y;j<m;j++)
            {
                if(a[i][j]=='X')
                {
                    vis[i][j]=1;
                    cnt++;
                }
                else break;
            }
            if(sum==-1)sum=cnt;
            else
            {
                if(sum==cnt)continue;
                else return 0;
            }
        }
        else break;
    }
    return 1;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++)
        {
            scanf("%s",a[i]);
        }
        int ok=1;
        int flag=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(a[i][j]=='X')
                {
                    flag=1;
                    int tmp=Slove(i,j);
                    if(tmp==0)ok=0;
                    break;
                }
            }
            if(flag==1)break;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(vis[i][j]==0&&a[i][j]=='X')ok=0;
            }
        }
        if(ok==1)printf("YES\n");
        else printf("NO\n");
    }
}