题目链接：CF 699B
题面：
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：
问能否在图中找到一个点，使得该点的十字爆炸，可以消除图中的所有墙（*）。
解题：
之前各种取巧，不是这错，就是那错。CF第2题，还是要往简单想，直接统计每行每列星数，然后枚举位置即可。
代码：
#include <iostream>
#include <cstdio>
using namespace std;
char mapp[1002][1002];
int rowc[1002];
int colc[1002];
int n,m;
int check(int x,int y)
{
int res=0;
res+=rowc[x];
res+=colc[y];
if(mapp[x][y]=='*')
res--;
return res;
}
int main()
{
int x,y,cnt=0,tmp;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
scanf(" %c",&mapp[i][j]);
if(mapp[i][j]=='*')
{
rowc[i]++;
colc[j]++;
cnt++;
}
}
}
bool flag=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
tmp=check(i,j);
if(tmp==cnt)
{
flag=1;
x=i;
y=j;
break;
}
}
if(flag)break;
}
if(flag)
printf("YES\n%d %d\n",x,y);
else
printf("NO\n");
return 0;
}