Codeforces #363(Div.2) B.One Bomb

最新推荐文章于 2019-01-30 23:28:29 发布

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本文介绍了一个基于矩形网格的游戏挑战，玩家需要放置一枚炸弹来消除所有障碍物。文章提供了问题描述、输入输出示例及解决方案代码，展示了如何通过计算行和列中障碍物的数量来确定可行的炸弹位置。

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 triggeringexactly 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

思路：暴力行列数相加等于总数或总数+1；不要用奇奇怪怪的方法！！不要用奇奇怪怪的方法！！不要用奇奇怪怪的方法！！！！！

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>

using namespace std;

char a[1000+5][1000+5];
int vish[1000 + 5],visl[1000 + 5];
int main()
{

    int n,m,sum = 0;
    cin>>n>>m;
    memset(vish,0,sizeof vish);
    memset(visl,0,sizeof visl);
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {

            cin>>a[i][j];
            if(a[i][j] == '*')
            {
                vish[i]++;
                visl[j]++;
                sum++;
            }
        }

    }

    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            if(a[i][j]=='*' && sum == vish[i]-1+visl[j])
            {
                cout<<"YES"<<endl<<i<<" "<<j<<endl;
                return 0;
            }
            else if(a[i][j] != '*' && sum == vish[i]+visl[j])
             {
                cout<<"YES"<<endl<<i<<" "<<j<<endl;
                return 0;
            }

    cout<<"NO"<<endl;
    return 0;
}