Window Pains
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 1090 |
|
Accepted: 540 |
Description
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 |
9 | |
When 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
1and 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
圖論的好題
把模型建為網(wǎng)絡(luò)，然后判斷是否為AOV網(wǎng)
如何構(gòu)圖
預(yù)處理要先計(jì)算出4*4格的位置可能填放的窗口
讀取快照后，對每一點(diǎn)處理如下
該點(diǎn)當(dāng)前的窗口為k，對該點(diǎn)可能出現(xiàn)窗口i,標(biāo)記g[k][i]有邊
正常的話，不會(huì)出現(xiàn)環(huán)
這里判斷AOV網(wǎng)用點(diǎn)的入度計(jì)算
如果存在超過未刪除的點(diǎn)的入度全部大于0，說明存在環(huán)
#include<algorithm>
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
string cover[4][4];
bool exist[10];//第i個(gè)窗口是否在快照中出現(xiàn)
int id[10];//入度
bool g[10][10];
int t;//頂點(diǎn)數(shù)
int n=4;
int a[4][4];
void getweizhi()
{
int i,j,k;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
cover[i][j].erase();
}
for(k=1;k<=9;k++)
{
i=(k-1)/3;
j=(k-1)%3;
cover[i][j]+=char(k+'0');
cover[i][j+1]+=char(k+'0');
cover[i+1][j]+=char(k+'0');
cover[i+1][j+1]+=char(k+'0');
}
}
void init()
{
int i,j,k;
memset(g,0,sizeof(g));
memset(exist,false,sizeof(exist));
memset(id,0,sizeof(id));
t=0;
for(i=0;i<4;i++)
{
for(j=0;j<4;j++)
{
cin>>k;
a[i][j]=k;
if (!exist[k])
{
t++;
exist[k]=true;
}
}
}
}
void build()
{
int i,j,p;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
for(p=0;p<=cover[i][j].length()-1;p++)
if (cover[i][j][p]-'0'!=a[i][j]&&!(g[a[i][j]][cover[i][j][p]-'0']))
{
g[a[i][j]][cover[i][j][p]-'0']=true;
id[cover[i][j][p]-'0']++;
}
}
}
bool check()
{
int i,j,k;
for(k=0;k<t;k++)
{
i=1;
while(!exist[i]||(i<=9&&id[i]>0)) i++;
if (i>9)//剩余的點(diǎn)中每個(gè)點(diǎn)入度都超過0
{
return false;
}
exist[i]=false;
for(j=1;j<=9;j++)
if (exist[j]&&g[i][j]) id[j]--;
}
return true;
}
int main()
{
string tmp;
getweizhi();
while(cin>>tmp)
{
if (tmp=="ENDOFINPUT")
{
break;
}
init();
build();
if (check())
cout<<"THESE WINDOWS ARE CLEAN"<<endl;
else
cout<<"THESE WINDOWS ARE BROKEN"<<endl;
cin>>tmp;
}
return 0;
}