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 . . .
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:
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.
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
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<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 #include<queue> 6 #include<vector> 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<<cnnt<<" "; 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 }