CodeForces745B B - Hongcow Solves A Puzzle 暴力+判断

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 n by 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.

Example

Input

2 3XXXXXX

Output

YES

Input

2 2.XXX

Output

NO

Input

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

Output

YES

Note

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

111222111222

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................

判断是否为正方形

#include <iostream> #include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define mod 1000000007#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  long long ll;typedef  unsigned long long  ull; inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int mapp[505][505];bool judge(int x,int y,int ct);int n,m,ct;int main(){ios::sync_with_stdio(false); cin >> n>>m;int endx=-1,endy=-1;for(int i=0;i<n;i++){string s;cin >>s;for(int j=0;j<s.size();j++){if(s[j]=='.'){mapp[i][j] = 0; }else{mapp[i][j] = 1;endx = i;endy = j; ct++;}} }/*for(int i=0;i<n;i++){for(int j=0;j<m;j++){cout << mapp[i][j]<<" ";}cout <<endl;}*/if(endx==-1){cout << "NO"<<endl;}else{if(judge(endx,endy,ct)){cout << "YES"<<endl;}else{cout << "NO"<<endl;}} return 0; }bool judge(int x,int y,int ct){int lenx=0,leny=0;for(int i=x;i>=0;i--){if(mapp[i][y]==1)lenx++; elsebreak;} for(int i=y;i>=0;i--){if(mapp[x][i]==1)leny++; elsebreak;} if(ct!=lenx*leny){return false;}//cout << lenx <<":"<<leny<<endl;for(int i=x;i>=x-lenx+1;i--){for(int j=y;j>=y-leny+1;j--){if(mapp[i][j]==0)return false;} } return true;}