Codeforces Round #363 (Div. 2) B 暴力

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B. One Bomb

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a description of a depot. It is a rectangular checkered field of n × m size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").

You have one bomb. If you lay the bomb at the cell (x, y), then after triggering it will wipe out all walls in the row x and all walls in the column y.

You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.

Input

The first line contains two positive integers n and m (1 ≤ n, m ≤ 1000) — the number of rows and columns in the depot field.

The next n lines contain m symbols "." and "*" each — the description of the field. j-th symbol in i-th of them stands for cell (i, j). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is
occupied by a wall.

Output

If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).

Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.

Examples

input

3 4

.*..

....

.*..

output

YES

1 2

input

3 3

..*

.*.

*..

output

NO

input

6 5

..*..

..*..

*****

..*..

..*..

..*..

output

YES

3 3

题意：

n*m的矩阵，'*'表示墙，'.'表示空地，现在在任意位置安放一颗炸弹，能否炸掉所有的墙，炸掉炸的范围是所在整行和所在整列

思路：

先统计出墙的总数，枚举炸的每个位置，如果整行整列的墙的个数等于总数，输出YES和位置，反之NO

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n,m;
string s[1010];
int a[1010][1010];
int r[1010],c[1010];
int main(){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) cin>>s[i];
for(int i=1;i<=n;i++){
for(int j=0;j<s[i].size();j++){
if(s[i][j]=='*') a[i][j+1]=1;
else a[i][j+1]=0;
}
}
int cnt=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(a[i][j]) {
cnt++;
r[i]++;
c[j]++;
}
}
}
if(cnt==0){
cout<<"YES"<<endl;
cout<<1<<" "<<1<<endl;
return 0;
}
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(a[i][j]==1){
if(r[i]+c[j]-1==cnt){
cout<<"YES"<<endl;
cout<<i<<" "<<j<<endl;
return 0;
}
}
else {
if(r[i]+c[j]==cnt){
cout<<"YES"<<endl;
cout<<i<<" "<<j<<endl;
return 0;
}
}
}
}
cout<<"NO"<<endl;
return 0;
}
/*
2 2
*.
.*

2 2
.*
*.
*/