Codeforces Round #192 (Div. 1) A. Purification 贪心

A. Purification

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/329/problem/A

Description

You are an adventurer currently journeying inside an evil temple. After defeating a couple of weak zombies, you arrived at a square room consisting of tiles forming an n × n grid. The rows are numbered 1 through n from top to bottom, and the columns are numbered 1 through n from left to right. At the far side of the room lies a door locked with evil magical forces. The following inscriptions are written on the door:

The cleaning of all evil will awaken the door!

Being a very senior adventurer, you immediately realize what this means. You notice that every single cell in the grid are initially evil. You should purify all of these cells.

The only method of tile purification known to you is by casting the "Purification" spell. You cast this spell on a single tile — then, all cells that are located in the same row and all cells that are located in the same column as the selected tile become purified (including the selected tile)! It is allowed to purify a cell more than once.

You would like to purify all n × n cells while minimizing the number of times you cast the "Purification" spell. This sounds very easy, but you just noticed that some tiles are particularly more evil than the other tiles. You cannot cast the "Purification" spell on those particularly more evil tiles, not even after they have been purified. They can still be purified if a cell sharing the same row or the same column gets selected by the "Purification" spell.

Please find some way to purify all the cells with the minimum number of spells cast. Print -1 if there is no such way.

Input

The first line will contain a single integer n (1 ≤ n ≤ 100). Then, n lines follows, each contains n characters. The j-th character in the i-th row represents the cell located at row i and column j. It will be the character 'E' if it is a particularly more evil cell, and '.' otherwise.

Output

If there exists no way to purify all the cells, output -1. Otherwise, if your solution casts x "Purification" spells (where x is the minimum possible number of spells), output x lines. Each line should consist of two integers denoting the row and column numbers of the cell on which you should cast the "Purification" spell.

Sample Input

3
.E.
E.E
.E.

Sample Output

1 1
2 2
3 3

HINT

题意

给你一个n*n的格子，你可以释放咒语，净化这一行和一列的格子

但是有一些非常邪恶的格子，你不能释放咒语，然后问你

最少释放多少次咒语，可以净化所有的格子

（注：每种格子都得净化，不管是E还是.

题解：

一定是n次，可以直接贪心的去找就好了，如果存在一行以及一列全是E的话，就输出-1

代码

#include<iostream>
#include<stdio.h>
using namespace std;

string s[205];
int main()
{
    int n;scanf("%d",&n);
    for(int i=0;i<n;i++)
        cin>>s[i];
    int flag1,flag2;
    flag1=flag2=0;
    for(int i=0;i<n;i++)
    {
        flag1=1;
        for(int j=0;j<n;j++)
        {
            if(s[i][j]=='.')
                flag1=0;
        }
        if(flag1)break;
    }
    for(int i=0;i<n;i++)
    {
        flag2=1;
        for(int j=0;j<n;j++)
        {
            if(s[j][i]=='.')
                flag2=0;
        }
        if(flag2)break;
    }
    if(flag1&&flag2)
    {
        printf("-1
");
        return 0;
    }
    if(!flag1)
    {
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(s[i][j]=='.')
                {
                    printf("%d %d
",i+1,j+1);
                    break;
                }
            }
        }
        return 0;
    }
    if(!flag2)
    {
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(s[j][i]=='.')
                {
                    printf("%d %d
",j+1,i+1);
                    break;
                }
            }
        }
    }
}