Problem 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.

Examples

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

Note

The first example is illustrated as follows. Purple tiles are evil tiles that have not yet been purified. Red tile is the tile on which "Purification" is cast. Yellow tiles are the tiles being purified as a result of the current "Purification" spell. Green tiles are tiles that have been purified previously.

In the second example, it is impossible to purify the cell located at row 1 and column 1.

For the third example:

题意：给出一张 n*n 的图，整张图需要净化，在图中 " . "代表可以净化的点，在其上可以净化这个点所在的行、列，" E " 代表无法去净化的点，即无法到达，只能通过 " . " 的净化来净化，如果能净化整张图，就输出每个格子的位置，否则输出 -1

思路：

看题意比较难，但观察下面的 Note 仔细想一想并没有那么难

首先，如同样例 2，对于某一行列均是 E 的情况，那么注定无解，因此这种情况直接输出 -1 即可

其次，若是某几行全是 E，那么为不与第一种情况冲突，可以保证剩下的行有足够的 " . " 使得可覆盖全图，对于这种情况，进行构造即可，每一列都输出第一个 " . " 的位置，可保证一定能覆盖全图

然后，若是某几列全是 E，那么为不与第一种情况冲突，可以保证剩下的列有足够的 " . " 使得可覆盖全图，对于这种情况，进行构造即可，每一行都输出第一个 " . " 的位置，可保证一定能覆盖全图

对于其他的情况，一定是行、列没有全是 E 的情况，那么同样可以找每一行每一列第一个 " . " 的位置即可

最后，根据以上 4 种情况，进行暴力搜索即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<deque>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define INF 0x3f3f3f3f
#define N 10001
#define LL long long
const int MOD=998244353;
const int dx[]={-1,1,0,0};
const int dy[]={0,0,-1,1};
using namespace std;
char G[N][N];
int row[N],col[N];//记录每一行每一列的E的个数
int main() {int n;cin>>n;for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)cin>>G[i][j];bool flagRow=false;bool flagCol=false;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(G[i][j]=='E'){row[i]++;//第i行E的个数+1col[j]++;//第j列E的个数+1if(row[i]==n)flagRow=true;if(col[j]==n)flagCol=true;if(flagRow&&flagCol){//某行某列E的个数全是E的情况cout<<-1<<endl;return 0;}}}}if(flagRow){//行全是E的情况for(int j=1;j<=n;j++){for(int i=1;i<=n;i++){if(G[i][j]=='.'){cout<<i<<" "<<j<<endl;break;}}}}else if(flagCol){//列全是E的情况for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(G[i][j]=='.'){cout<<i<<" "<<j<<endl;break;}}}}else{//行或列不存在全是E的情况for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(G[i][j]=='.'){cout<<i<<" "<<j<<endl;break;}}}}return 0;
}

Purification（CF-330C）相关推荐

【解题报告】随便练练二（CF 2300）
[解题报告]随便练练二(CF 2300) A:Antimatter | CF383D 题意思路 :DP 代码 B:Physical Education Lessons | CF915E 题意思路一 ...
Commentator problem（CF 2）
题目链接题目大意: 给定三个圆,询问是否存在点满足该点与三个圆夹角均相等,若存在多组解返回夹角最大值. 圆外一点到两圆夹角均相等: 即 sina = sinb = r1 / d1 = r2 / d2 ...
D. Make a Power of Two（cf#739DIV3）
D. Make a Power of Two 链接: link. 题意: 找出将数字转换为 2 的任意幂的最小移动次数. 题解: 先将2的xxx次幂的结果以字符串形式保存,输入字符串nnn后,因为存在 ...
Web of Lies（CF 1548A）
这是今天在打个人赛时碰见的一道题,是一道半图论半思维的题. Web of Lies 题目大意不难理解,在这里只需要注意一些细节.在加边时,只有当cnt[min]的值为1时答案才应该减1,而不是当cnt ...
Magic Powder - 2 （CF 670_D）
http://codeforces.com/problemset/problem/670/D2 The term of this problem is the same as the previous ...
【解题报告】博弈专场（CF 2000~2200）前五题
[解题报告]博弈专场 (CF 2000+)前五题 A:Fox and Card Game | CF388C 题意思路代码 B:Berzerk | CF786A 题意思路代码 C:Ithea P ...
软件设计师提纲+复习资料整理（上午题）
文章目录软件设计师考试大纲上午题(选择题) 一.计算机组成原理考点:CPU结构组成考点:原码.反码.补码定点整数范围考点:浮点数表示考点:RISC和CISC计算机的区别考点:奇校验与偶校 ...
c性能大容量cket_5千左右预算，既轻薄（高颜值）又高性能的笔记本推荐（畅玩LOL、CF、DNF、流放之路、梦幻西游）...
在目前笔记本制造技术中,轻薄和性能二者不可兼得,轻薄的本性能不高,高性能的太厚重, 这里推荐一些既轻薄,外观好看,同时又高性能的轻薄本. 很多轻薄本多为低压U+集成显卡,性能太弱,都不适合玩大型3D网 ...
基于矩阵分解的CF算法实现（一）：（Funk SVD）LFM
基于矩阵分解的CF算法实现(一):LFM LFM也就是前面提到的Funk SVD矩阵分解 LFM原理解析 LFM(latent factor model)隐语义模型核心思想是通过隐含特征联系用户和物品 ...
2021.5.10（cf）
1.cf D. Maximum Sum of Products https://codeforces.com/contest/1519/problem/D 大意:给定一个数字n代表序列的长度,接下来n ...

Purification（CF-330C）

Problem Description

Input

Output

Examples

Note

Source Program

Purification（CF-330C）相关推荐

最新文章

热门文章