codeforces 699B One Bomb

来源：互联网发布：模拟人生4男性捏脸数据编辑：程序博客网时间：2024/05/16 19:08

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
YES1 2
input
3 3..*.*.*..
output
NO
input
6 5..*....*..*****..*....*....*..
output
YES3 3

题意：*表示墙壁，一颗炸弹可以炸掉与它同一行，同一列的所有墙壁。问只用一颗炸弹是否能图中所有的墙壁炸毁，若能，给出炸弹任意一个能安放的位置？

题解：记录每一行，每一列的墙壁数，然后组合所有的行与列，代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char map[1010][1010];int row[1010],line[1010];int main(){int n,m,i,j,num;while(scanf("%d%d",&n,&m)!=EOF){memset(row,0,sizeof(row));memset(line,0,sizeof(line));num=0;for(i=0;i<n;++i){scanf("%s",map[i]);for(j=0;j<m;++j){if(map[i][j]=='*'){num++;  row[i]++;  line[j]++;}}}int sign=0;for(i=0;i<n;++i){for(j=0;j<m;++j){int cnt=row[i]+line[j];if(map[i][j]=='*')cnt--;if(num==cnt){printf("YES\n%d %d\n",i+1,j+1);sign=1;break;}}if(sign)break;}if(!sign)printf("NO\n");}return 0;}

0 0