A. Purification
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 1through 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 test(s)
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:

这题是NOI 2013 被虐残以后 做的第一次Div 1,马上就掉回了轻囧

这题一开始以为DLX各种弱。。。。。

印度尼西亚的人怎么都出这种游戏题目、、、、

进入正解模式:

我们发现出现‘十‘E 是无解的

否则必然存在n次Purification(净化)的解.

反证:

假设最优解为n+1

那么必然有一行/一列净化了2次,删掉那次不影响结果

假设存在答案为n的净化方案:

因为只做了n遍,必然每行取一个(%号表示净化点)

%E

%E

或每列:

%%

EE

所以考虑这2种情况,随便取即可,反之无解

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,a[MAXN]={0},d[MAXN]={0};
bool b[MAXN][MAXN]={0};
int main()
{
// freopen("Purification.in","r",stdin);
cin>>n;
For(i,n) For(j,n)
{
char c;
while (cin>>c)
{
if (c=='.') b[i][j]=1,a[i]=j,d[j]=i;
else if (c=='E') b[i][j]=0;
else continue;
break;
}
}
/*
For(i,n)
{
bool bo=0;
For(j,n) if (b[i][j]) bo=1;
if (!bo) {puts("-1");return 0;}
bo=0;
For(j,n) if (b[j][i]) bo=1;
if (!bo) {puts("-1");return 0;}
}*/
//For(i,n) cout<<a[i]<<' ';cout<<endl;
//For(i,n) cout<<d[i]<<' ';cout<<endl; bool bo=0;
For(i,n) if (!a[i]) bo=1;
if (!bo)
{
For(i,n) cout<<i<<' '<<a[i]<<endl;
return 0;
} bo=0;
For(i,n) if (!d[i]) bo=1;
if (!bo)
{
For(i,n) cout<<d[i]<<' '<<i<<endl;
return 0;
} puts("-1"); return 0;
}

CF 329A(Purification-贪心-非DLX)的更多相关文章

  1. 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/probl ...

  2. CF 949D Curfew——贪心(思路!!!)

    题目:http://codeforces.com/contest/949/problem/D 有二分答案的思路. 如果二分了一个答案,首先可知越靠中间的应该大约越容易满足,因为方便把别的房间的人聚集过 ...

  3. CF Covered Path (贪心)

    Covered Path time limit per test 1 second memory limit per test 256 megabytes input standard input o ...

  4. CF 389 E 贪心(第一次遇到这么水的E)

    http://codeforces.com/contest/389/problem/E 这道题目刚开始想的特别麻烦...但是没想到竟然是贪心 我们只需要知道偶数的时候可以对称取的,然后奇数的时候没次取 ...

  5. CF 463A && 463B 贪心 && 463C 霍夫曼树 && 463D 树形dp && 463E 线段树

    http://codeforces.com/contest/462 A:Appleman and Easy Task 要求是否全部的字符都挨着偶数个'o' #include <cstdio> ...

  6. cf 之lis+贪心+思维+并查集

    https://codeforces.com/contest/1257/problem/E 题意:有三个集合集合里面的数字可以随意变换位置,不同集合的数字,如从第一个A集合取一个数字到B集合那操作数+ ...

  7. cf C. Purification

    http://codeforces.com/contest/330/problem/C 这道题分三种情况.有一行全是E,有一列全是E,还有一种为无解的情况. #include <cstdio&g ...

  8. CF# 260 A. Laptops

    One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the m ...

  9. Codeforces Round #401 (Div. 2) 离翻身就差2分钟

    Codeforces Round #401 (Div. 2) 很happy,现场榜很happy,完全将昨晚的不悦忘了.终判我校一片惨白,小董同学怒怼D\E,离AK就差一个C了,于是我AC了C题还剩35 ...

随机推荐

  1. Python 项目实践二(生成数据)第二篇

    接着上节继续学习,在本节中,我们将使用Python来生成随机漫步数据,再使用matplotlib以引人瞩目的方式将这些数据呈现出来.随机漫步是这样行走得到的路径:每次行走都完全是随机的,没有明确的方向 ...

  2. SOA和Web Servcie的区别

    soa(Service-Oriented Architecture)是服务对服务的,web service是服务对客户端的. 都提供服务.  服务接口都是基于开发的.  服务接口和服务的具体实现都是分 ...

  3. 负载均衡介绍及Nginx简单实现

    负载均衡介绍及Nginx简单实现 负载均衡 负载均衡介绍及Nginx简单实现 1. 介绍 2. 常用的开源软件 2.1 LVS 优点 缺点 2.2 Nginx 优点 缺点 3. 常用的开源反向代理软件 ...

  4. 喵哈哈村的魔法考试 Round #3 (Div.2) 题解

    A 题解:保证一个三角形的话,得两边之和大于第三边才行,所以都拿来判一判就好了. #include <iostream> using namespace std; int main(){ ...

  5. UVALive 6916 Punching Robot dp

    Punching Robot 题目连接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid= ...

  6. Hadoop系列之(三):使用Cloudera部署,管理Hadoop集群

    1. Cloudera介绍 Hadoop是一个开源项目,Cloudera对Hadoop进行了商业化,简化了安装过程,并对hadoop做了一些封装. 根据使用的需要,Hadoop集群要安装很多的组件,一 ...

  7. 并发编程(二):全视角解析volatile

    一.目录 1.引入话题-发散思考 2.volatile深度解析 3.解决volatile原子性问题 4.volatile应用场景 二.引入话题-发散思考 public class T1 { /*vol ...

  8. spring cloud 学习(11) - 用fastson替换jackson及用gb2312码输出

    前几天遇到一个需求,因为要兼容旧项目的编码格式,需要spring-cloud的rest接口,输出gb2312编码,本以为是一个很容易的事情,比如下面这样: @RequestMapping(method ...

  9. ChibiOS/RT 2.6.9 CAN Driver

    Detailed Description Generic CAN Driver. This module implements a generic CAN (Controller Area Netwo ...

  10. ubuntu的配置文件

    ubuntu的配置文件 是 ~/.gconf 我是把终端弄挂了, 只能再桌面系统下找到 ~/.gconf 下的相应文件 修改后就恢复到原来状态.