[POJ 2585] Window Pains【拓扑判环】

  • 2018-01-04
  • 0
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Problem:

Time Limit: 1000MS Memory Limit: 65536K

Description

Boudreaux likes to multitask, especially when it comes to using his computer. Never satisfied with just running one application at a time, he usually runs nine applications, each in its own window. Due to limited screen real estate, he overlaps these windows and brings whatever window he currently needs to work with to the foreground. If his screen were a 4 x 4 grid of squares, each of Boudreaux's windows would be represented by the following 2 x 2 windows:

1 1 . .
1 1 . .
. . . .
. . . .
. 2 2 .
. 2 2 .
. . . .
. . . .
. . 3 3
. . 3 3
. . . .
. . . .
. . . .
4 4 . .
4 4 . .
. . . .
. . . .
. 5 5 .
. 5 5 .
. . . .
. . . .
. . 6 6
. . 6 6
. . . .
. . . .
. . . .
7 7 . .
7 7 . .
. . . .
. . . .
. 8 8 .
. 8 8 .
. . . .
. . . .
. . 9 9
. . 9 9

When Boudreaux brings a window to the foreground, all of its squares come to the top, overlapping any squares it shares with other windows. For example, if window 1and then window 2 were brought to the foreground, the resulting representation would be:

1 2 2 ?
1 2 2 ?
? ? ? ?
? ? ? ?
If window 4 were then brought to the foreground:
1 2 2 ?
4 4 2 ?
4 4 ? ?
? ? ? ?

. . . and so on . . .
Unfortunately, Boudreaux's computer is very unreliable and crashes often. He could easily tell if a crash occurred by looking at the windows and seeing a graphical representation that should not occur if windows were being brought to the foreground correctly. And this is where you come in . . .

Input

Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.

A single data set has 3 components:

  1. Start line - A single line:
    START
  2. Screen Shot - Four lines that represent the current graphical representation of the windows on Boudreaux's screen. Each position in this 4 x 4 matrix will represent the current piece of window showing in each square. To make input easier, the list of numbers on each line will be delimited by a single space.
  3. End line - A single line:
    END

After the last data set, there will be a single line:
ENDOFINPUT

Note that each piece of visible window will appear only in screen areas where the window could appear when brought to the front. For instance, a 1 can only appear in the top left quadrant.

Output

For each data set, there will be exactly one line of output. If there exists a sequence of bringing windows to the foreground that would result in the graphical representation of the windows on Boudreaux's screen, the output will be a single line with the statement:

THESE WINDOWS ARE CLEAN

Otherwise, the output will be a single line with the statement:
THESE WINDOWS ARE BROKEN

Sample Input

START
1 2 3 3
4 5 6 6
7 8 9 9
7 8 9 9
END
START
1 1 3 3
4 1 3 3
7 7 9 9
7 7 9 9
END
ENDOFINPUT

Sample Output

THESE WINDOWS ARE CLEAN
THESE WINDOWS ARE BROKEN

Source

South Central USA 2003

Solution:

很好的拓扑排序题,有些同学把它当做大暴力来做真是可惜了(数据都不出强一点,卡掉暴力多好~~)。

若每个 2 × 2 块 id 应该出现的位置被另一个块 cov 占据,则添加一条 cov -> id 的边,表示在完整看到 id 前必须将 cov 先“关闭”

而如果电脑系统没有崩溃,总有一种关闭窗口的方法使得所有窗口在关闭时都能被完整看到

这样问题就被转化为判断 AOV 图上是否有环,可以用拓扑排序完成。若没有入度为 0 的点时还有点未被加入拓扑序列,则图上存在环。

Code: O(T(N2+M)), N为屏幕尺寸, M为窗口总数 [160K, 0MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

char op[15];
int scr[4][4], indeg[10];
bool ed[10][10];
int deg0[10], topd, cnt;

int main(){
	while(scanf("%s", op) != EOF && op[0] == 'S'){
		for(register int i = 0; i < 4; i++)
			for(register int j = 0; j < 4; j++)
				scanf("%d", &scr[i][j]);
		scanf("%s", op);  // Skip "END"
		memset(ed, 0, sizeof(ed)), memset(indeg, 0, sizeof(indeg)), topd = cnt = 0;
		for(register int i = 0; i < 3; i++)
			for(register int j = 0; j < 3; j++){
				int id = i * 3 + j + 1, cov = scr[i][j];
				if(cov != id && !ed[cov][id]) ed[cov][id] = 1, indeg[id]++;
				cov = scr[i + 1][j];
				if(cov != id && !ed[cov][id]) ed[cov][id] = 1, indeg[id]++;
				cov = scr[i][j + 1];
				if(cov != id && !ed[cov][id]) ed[cov][id] = 1, indeg[id]++;
				cov = scr[i + 1][j + 1];
				if(cov != id && !ed[cov][id]) ed[cov][id] = 1, indeg[id]++;
				if(!indeg[id]) deg0[++topd] = id;
			}
		while(topd){
			int u = deg0[topd--];
			cnt++;
			for(register int v = 1; v <= 9; v++)
				if(ed[u][v]) if(!--indeg[v]) deg0[++topd] = v;
		}
		if(cnt < 9) puts("THESE WINDOWS ARE BROKEN");
		else puts("THESE WINDOWS ARE CLEAN");
	}
	return 0;
}

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OPEN AT 2017.12.10

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