codeforces 699 B. One Bomb (思维)

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

题意：如上矩阵，问只放一颗炸弹是否能将全部的墙(图中'*')炸烂,并输出炸弹的坐标
思路：题中说了炸弹可以炸烂所在的同行同列的所有墙，那么遍历n*m个位置考虑这个位置所在的同行同列的所有墙是否等于总墙数，
用x[],y[]分别表示每行和每列的墙数，然后比较x[i]+y[j]-1=?num
注意：（1）矩阵可能没有'*',所以直接输出坐标1 1 ;
（2）炸弹位置可能在'.',比如：
.*
*.
1 1 位置是可行方案
代码:

#include<cstdio>
#include<cstring>
char map[1010][1010];
int main()
{
int n,m;
int x[1010],y[1010],cnt=0;
memset(x,0,sizeof(x));
memset(y,0,sizeof(y));
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
scanf("%s",map[i]);
for(int j=0;j<m;j++)
{
if(map[i][j]=='*')
{
cnt++;
x[i]++;
y[j]++;
}
}
}
int flag=0,px=1,py=1;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(map[i][j]=='*')
{
if(x[i]+y[j]-1==cnt)
{
flag=1;
px=i+1;
py=j+1;
break;
}
}
if(map[i][j]=='.')
{
if(x[i]+y[j]==cnt)
{
flag=1;
px=i+1;
py=j+1;
break;
}
}
}
if(flag)
break;
}
if(flag||cnt==0)
printf("YES\n%d %d\n",px,py);
else
printf("NO\n");
return 0;
}