Codeforces #363（Div.2）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

题目大意：给你一个n*m的图，其中星号表示要炸的区域，给你一个炸弹，这个炸弹能够炸当前坐标对应的行和列上的星号，问一个炸弹是否够用，如果够用，输出对应坐标。

思路：

1、统计星号的个数cont

2、统计每一行、每一列上边星号的个数。

3、然后对坐标每一个点都进行枚举，对应当前坐标能够搞定的星号个数为：tmp=row【i】+col【j】；if（a【i】【j】是星号，对应值减一.）如果tmp==cont，那么当前坐标就是可行解。

4、注意没有星号的情况。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
char a[1005][1005];
int col[1005];
int row[1056];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int ok=1;
for(int i=0;i<n;i++)
{
scanf("%s",a[i]);
}
int cont=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(a[i][j]=='*')
{
col[j]++;
row[i]++;
cont++;
}
}
}
if(cont==0)
{
printf("YES\n 1 1\n");
continue;
}
int x,y;
int flag=0;
x=-1;y=-1;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
int tmp=row[i]+col[j];
if(a[i][j]=='*')tmp--;
if(tmp==cont)
{
flag=1;
x=i;y=j;
break;
}
}
if(flag==1)break;
}
if(flag==1)
{
printf("YES\n");
printf("%d %d\n",x+1,y+1);
}
else printf("NO\n");
}
}