【POJ 2585】Window Pains 拓扑排序

Description

. . . 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
-----------------------------------------------------------------------------
最早想到的是dfs 从最上面一层倒着往回推，但最后没法记录点了.....

其实就是个拓扑排序的水题.....
主要是建边：枚举9个炮台，如果在自己射程区域内不是自己，说明被覆盖，因此说明这个颜色是当前颜色的先决条件，
于是连一条边。（开始理解为只要不是自己就连边，结果根本没有入度为0的边23333
然后跑一遍拓扑排序，如果没成环就成功了，成环就失败。
这是代码：

 #include<iostream>

 #include<cstdio>

 #include<algorithm>

 #include<cmath>

 #include<queue>

 #include<vector>

 #define N 20

 using namespace std;

 struct node

 {

     int u,v,nxt;

 }e[N*];

 int first[N],cnt;

 void ade(int u,int v)

 {

     e[++cnt].nxt=first[u]; first[u]=cnt;

     e[cnt].u=u; e[cnt].v=v;

 }

 int ru[N],cnnt;

 int dir[][]= {,, ,, ,, ,};

 int local[][]= {-,-, ,, ,, ,, ,, ,, ,, ,, ,, ,};

 void topsort()

 {

     priority_queue<int>q;

     for(int i=;i<=;i++)

         if(ru[i]==) q.push(i);

     while(!q.empty())

     {

         int x=q.top(); q.pop();

         ++cnnt;

     //    cout<<cnnt<<" ";

         for(int i=first[x];i;i=e[i].nxt)

         {

             int v=e[i].v;

             ru[v]--;

             if(ru[v]==) q.push(v);

         }

     }

     if(cnnt!=) printf("THESE WINDOWS ARE BROKEN\n");

     else cout<<"THESE WINDOWS ARE CLEAN"<<endl;

 }

 int a[][];

 bool vis[N][N];

 int main()

 {

     char str[];

     while(scanf("%s",str),strcmp(str,"ENDOFINPUT"))

     {

         for(int i=;i<;i++)

             for(int j=;j<;j++)

                 scanf("%d",&a[i][j]);

         scanf("%s",str);

         memset(e,,sizeof(e));

         memset(vis,,sizeof(vis));

         memset(first,,sizeof(first));

         memset(ru,,sizeof(ru));

         cnt=cnnt=;

         for(int k=;k<=;k++)

             for(int i=;i<;i++)//在自己射程区域内不是自己->被覆盖

             {

                 int x=local[k][]+dir[i][];

                 int y=local[k][]+dir[i][];

                 int now=a[x][y];

                 if(k!=now&&!vis[k][now])

                 {

                     vis[k][now]=;

                     ade(k,now);

                     ru[now]++;

                 }

             }

         topsort();

     }

     return ;

 }