【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: 

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
-----------------------------------------------------------------------------
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 ;
 }