poj 2585 Window Pains（拓扑结构）

Window Pains

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 9When 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 1 and 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:
Start line - A single line:
START

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.
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

以下思路来源于链接

思路：
嗯，这题能建个有向图，荧屏上的数字出现后出现的会把先出现的覆盖，就形成了约束条件。
建图：按左上角开始每个数字的首个数字出现的位置用数组记录，遍历1～9九个数字的相邻四个位置的数字是否==记录首个数字，等于的话就没被覆盖，不等于就是被覆盖了，可以建一条原记录数字到当前显示数字的一条有向边。

Code：

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
using namespace std;

int d[10][2]={-1,-1, 0,0, 0,1, 0,2, 1,0, 1,1, 1,2, 2,0, 2,1, 2,2};//前面两个-1是凑数用的，没有用，剩下的就是遍历那九个数对应的起始点。 
int dd[4][2]={0,0, 0,1 ,1,1 ,1,0};//每一个起始点对应的那四个点 
int tu[15][15];//存数用的 
int in[15];//拓扑结构记录入度用的 
int bj[15][15];//存边用的 
int main()
{
	char s1[20],s2[20];
	int x,y,z;
	while(~scanf("%s",s1))
	{
		if(strcmp(s1,"ENDOFINPUT")==0)break;//结束条件 
		memset(bj,0,sizeof bj);//多组输入 
		memset(in,0,sizeof in);//进行初始化 
		for(int i=0;i<4;i++)
			for(int j=0;j<4;j++)
				scanf("%d",&tu[i][j]);//输入 
		scanf("%s",s2);
		for(int i=1;i<=9;i++)
			for(int j=0;j<4;j++)
			{
				x=d[i][0]+dd[j][0];//进行遍历每一个点的那四个点 
				y=d[i][1]+dd[j][1];
				z=tu[x][y];
				if(z!=i&&bj[i][z]==0){//如果不等于的话，就说明被覆盖了 
					bj[i][z]=1;//建立边 
					in[z]++;//入度+1 
				}
			}
		int ans=0;//初始化为0，用于记录 
		int k; 
		for(int i=1;i<=9;i++)
		{//遍历九个节点 
			for(int j=1;j<=9;j++){
				if(in[j]==0){//找到树根 
					k=j;//记录根节点，下面用于删除它所连的节点 
					ans++;//用于后期判断 
					break;
				}
			}
			in[k]--;//相当于删除操作，上面if以后不会在遍历到了 
			for(int j=1;j<=9;j++)
				if(bj[k][j])in[j]--;//删除该根对应的节点 
		}
		//如果ans为9的话说明可以形成一个无环图，否则为环了 
		 if(ans==9)printf("THESE WINDOWS ARE CLEAN\n"); //不为环则可以形成 
		 else printf("THESE WINDOWS ARE BROKEN\n");//为环则是程序崩溃了 
	}
	return 0;
}