CF330 C. Purification 认真想后就成水题了

C. Purification
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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 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 3Note
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:

$\$

题解：
题目就别看它描述了，直接看下面的样例解释就好了。就是E是不能放清除点的地方。
题解：
跟330A那题比较类似，如果你要清除一个n*n的正方形，你知道只用放n个点就能清除所有的方块。
如
*****
.....
.....
.....
.....
*的就是清除他们的点。同理竖着的，横着的，斜着的。所以我们只用构造出这样的5个点就可以了。所以题目转换成判断是否存在这样的n个点，以及如何放的问题。
第一类：
EEEEE
E....
E....
E....
E....
这是无解的情况，因为（1，1）这个点无法清除。所以判断无解的情况只要检索是否存在一行都是E并且一列都是E这种就OK了。
第2类：
EEEEE
E....
E.E..
E..E.
.....
像这种，横着存在全是E情况的，只用在每竖行找到一个能放置的点就可以了（一定存在的，不然就是上面所说的无解情况了）。
第3类：
EEEE.
E....
E.E..
E.E..
E....
这种类似于第2类分析。每个横行找到一个能放置的点就可以了（一定存在的，不然就是上面所说的无解情况了）。
如果是第4类这样的
EEEE.
E....
E....
E....
.....
可以直接用第2类的构造方法来做。所以这样就成了水题一道了。

/*
 * @author ipqhjjybj
 * @date  20130720
 *
 */
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std;
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define clr(x,k) memset(x,k,sizeof(x))

int col[200];//列
int row[200];//行

char s[105][105];
int r,c;
int main(){
    //freopen("330C.in","r",stdin);
    int n;
    scanf("%d",&n);
    r=c=0;
    getchar();
    for(int i=1;i <= n;i++)
        gets(s[i]+1);
    for(int i=1;i<=n;i++)
        for(int j = 1;j<=n;j++){
            if(s[i][j]=='E'){
                row[i]++,col[j]++;
                if(row[i]==n) r=1;
                if(col[j]==n) c=1;
                if(r&&c){
                    puts("-1");
                    return 0;
                }
            }
        }
    if(r)
        for(int j=1;j<=n;j++){
            for(int i = 1;i<=n;i++){
                if(s[i][j]=='.'){
                    printf("%d %d\n",i,j);
                    break;
                }
            }
        }
    else{
        for(int i = 1;i <= n;i++)
            for(int j = 1;j <= n;j++)
                if(s[i][j]=='.'){
                    printf("%d %d\n",i,j);
                    break;
                }
    }
    return 0;
}

补充：软件开发 , C++ ,