Codeforces 699B - One Bomb

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
预处理每一行每一列的墙数，最后暴力枚举炸弹位置就行了。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char a[1005][1005];int b[1005],c[1005];int main(){    int n,m;    memset(b,0,sizeof(b));    memset(c,0,sizeof(c));    scanf("%d%d",&n,&m);    for (int i=0;i<n;i++)    {        scanf("%s",a[i]);    }    int s=0;    for (int i=0;i<n;i++)    {        for (int j=0;j<m;j++)        {            if (a[i][j]=='*')            {              s++;              b[i]++;              c[j]++;            }        }    }    for (int i=0;i<n;i++)    {        for (int j=0;j<m;j++)        {            if (a[i][j]=='*')            {                if ((s-b[i]-c[j]+1)==0)                {                    printf("YES\n");                    printf("%d %d",i+1,j+1);                    return 0;                }            }else            {                if ((s-b[i]-c[j])==0)                {                    printf("YES\n");                    printf("%d %d",i+1,j+1);                    return 0;                }            }        }    }    printf("NO");}