分類： 貪心 棋牌方塊模擬 2013-07-21 14:33 103人閱讀 評論(0) 收藏 舉報
A. 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 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 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:
這題是NOI 2013 被虐殘以後 做的第一次Div 1，馬上就掉回藍了輕囧
這題一開始以為DLX各種弱。。。。。
印度尼西亞的人怎麼都出這種游戲題目、、、、
進入正解模式：
我們發現出現‘十‘E 是無解的
否則必然存在n次Purification（淨化）的解.
反證：
假設最優解為n+1
那麼必然有一行/一列淨化了2次，刪掉那次不影響結果
假設存在答案為n的淨化方案：
因為只做了n遍，必然每行取一個(%號表示淨化點)
%E
%E
或每列：
%%
EE
所以考慮這2種情況，隨便取即可，反之無解
#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;
}
