CodeForces 699B - One Bomb(思维)

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

(这里我用 ‘@’代替 ‘*’ ,因为在MarkDown的语法上会有一些问题)

题意:

给出一个地图 ‘.’代表路,’*’代表墙.问你能否在一个点上放一个炸弹,使得所有的墙都炸掉(炸弹往四个方向炸,威力无限,而且可以放在墙上).如果可以,输出YES和坐标,否则输出NO.

解题思路:

用两个数组记录下来每一列和每一行墙的个数.然后O(n^2)去试探每一个点.看看x[i]+y[i]和sum是不是相等(注意,如果map[i][j]是墙,那么要减去一个).

AC代码:

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char qdu[1000][1000];
int main()
{
int n;
int m;
scanf("%d%d",&n,&m);
getchar();
for(int i = 0;i < n;i++)    gets(qdu[i]);
//for(int i = 0;i < n;i++)    puts(qdu[i]);
int x[1000] = {0};
int y[1000] = {0};
int cnt = 0;
for(int i = 0;i < n;i++)
{
for(int j = 0;j < m;j++)
{
if(qdu[i][j] == '*')
cnt++;
}
}
for(int i = 0;i < n;i++)
{
for(int j = 0;j < m;j++)
{
if(qdu[i][j] == '*')
{
//printf("%d %d\n",i,j);
x[i]++;
y[j]++;
}
}
}
for(int i = 0;i < n;i++)
{
for(int j = 0;j < m;j++)
{
int res = x[i]+y[j];
if(qdu[i][j] == '*')
res--;
if(res == cnt)
{
printf("YES\n%d %d\n",i+1,j+1);
return 0;
}
}
}
printf("NO\n");
return 0;
}