作者：来杯冰柠檬水 | 来源：互联网 | 2023-08-17 08:21
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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：
1 #include<iostream>
2 #include
3 #include
4 #include
5 #include
6 #include
7 #define N 20
8 using namespace std;
9 struct node
10 {
11 int u,v,nxt;
12 }e[N*2];
13 int first[N],cnt;
14 void ade(int u,int v)
15 {
16 e[++cnt].nxt=first[u]; first[u]=cnt;
17 e[cnt].u=u; e[cnt].v=v;
18 }
19 int ru[N],cnnt;
20 int dir[4][2]= {0,0, 1,0, 0,1, 1,1};
21 int local[10][2]= {-1,-1, 0,0, 0,1, 0,2, 1,0, 1,1, 1,2, 2,0, 2,1, 2,2};
22 void topsort()
23 {
24 priority_queue<int>q;
25 for(int i=1;i<=9;i++)
26 if(ru[i]==0) q.push(i);
27 while(!q.empty())
28 {
29 int x=q.top(); q.pop();
30 ++cnnt;
31 // cout<
32 for(int i=first[x];i;i=e[i].nxt)
33 {
34 int v=e[i].v;
35 ru[v]--;
36 if(ru[v]==0) q.push(v);
37 }
38 }
39 if(cnnt!=9) printf("THESE WINDOWS ARE BROKEN
");
40 else cout<<"THESE WINDOWS ARE CLEAN"<<endl;
41 }
42 int a[5][5];
43 bool vis[N][N];
44 int main()
45 {
46 char str[20];
47 while(scanf("%s",str),strcmp(str,"ENDOFINPUT"))
48 {
49 for(int i=0;i<4;i++)
50 for(int j=0;j<4;j++)
51 scanf("%d",&a[i][j]);
52 scanf("%s",str);
53 memset(e,0,sizeof(e));
54 memset(vis,0,sizeof(vis));
55 memset(first,0,sizeof(first));
56 memset(ru,0,sizeof(ru));
57 cnt=cnnt=0;
58 for(int k=1;k<=9;k++)
59 for(int i=0;i<4;i++)//在自己射程区域内不是自己->被覆盖
60 {
61 int x=local[k][0]+dir[i][0];
62 int y=local[k][1]+dir[i][1];
63 int now=a[x][y];
64 if(k!=now&&!vis[k][now])
65 {
66 vis[k][now]=1;
67 ade(k,now);
68 ru[now]++;
69 }
70 }
71 topsort();
72 }
73 return 0;
74 }
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