codeforces 330c

在一个n×n的方格中,部分格子被标记为特别邪恶,不能直接净化。任务是在最少的操作次数下净化所有格子,否则输出-1。本文提供了一种解决方案,并附带源代码。

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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 numbered1 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 thei-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

Input
3
.E.
E.E
.E.
Output
1 1
2 2
3 3
Input
3
EEE
E..
E.E
Output
-1
Input
5
EE.EE
E.EE.
E...E
.EE.E
EE.EE
Output
3 3
1 3
2 2
4 4

5 3

分析:

首先是读懂题意。假使敢于尝试就可以做出来。

我的思路是这样的,现在n^2的时间标记行列的可消除情况,同时给行记数cnt,然后在n^2的时间内检查是否输出"-1",如果cnt大小为n,就按行输出否则,按列输出。

代码:

#include
#include
#include
using namespace std;
char maze[200][200];
int r[120],c[120];
int main()
{
    int n;

    while(~scanf("%d",&n))
    {
        for(int i=0;i

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