CF-One Bomb（炸弹问题）

B. One Bomb

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard 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

大致题意：n*m列的矩阵，*代表炸弹，试问是否存在一种情况使得只安放一个炸弹，使得所有炸弹全部消失。

思路：枚举每一个点，使得该点的行和列的炸弹消失，判断是否能消灭所有炸弹。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int visit1[1005],visit2[1005];
char box[1005][1005],str[1005][1005];
int  main()
{
int n,m,i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
int flag1=0,flag2=0;
int sum1=0,sum2;
int ykc1[1005],ykc2[1005];
int x=0,y=0;
memset(ykc1,0,sizeof(ykc1));
memset(ykc2,0,sizeof(ykc2));
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
scanf(" %c",&box[i][j]);
if(box[i][j]=='*')
{
sum1+=1;
ykc1[i]+=1;
ykc2[j]+=1;
}
}
}
int t1=0,t2=0;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{

sum2=ykc1[i]+ykc2[j];
if(box[i][j]=='*')
sum2-=1;
if(sum2==sum1)
{
flag1=1;
t1=i;
t2=j;
}
if(flag1==1)
break;
}
if(flag1==1)
break;
}
if(flag1==0)
printf("NO\n");
else
{
printf("YES\n");
printf("%d %d\n",t1,t2);
}
}
}