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Window Pains（点击>>原POJ)
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 1980 | Accepted: 998 |
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:
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 1 and then window 2 were brought to the foreground, the resulting representation would be:
. . . 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 . . .
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| If window 4 were then brought to the foreground: |
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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
Source
South Central USA 2003
i题意：
显示颜色的问题，一种颜色有一种显示的区域，大的可以覆盖小的。给定几个矩阵，问显示效果怎么样，好的话输出一句，否则输出另外一句。
具体参考书籍《图论》哈工大出版社。
i思路：
每个显色的数字位置如图：
这可是学姐在黑板上一点点画出来的图（这是我用EXCEL打的），讲题挺认真的，负责。
i参考代码：
实现方式：二维数组
#include<stdio.h>
#include<string.h>
const int MYDD=1103;int indegree[MYDD];
int map[113][113];
int dx[]= {0,1,0,1};
int dy[]= {0,0,1,1}; //移动的方向,注意不同于搜索
int local[10][2]= {-1,-1, 0,0, 0,1, 0,2, 1,0, 1,1, 1,2, 2,0, 2,1, 2,2};
//窗口的固定位置
bool TopoSort() {int now,flag;// now 当前选中的节点；flag 标记合法性for(int j=1; j<=9; j++) {flag=0;for(int i=1; i<=9; i++) {if(!indegree[i]) {flag=1;now=i;break;//入度为 0 即前驱}}if(!flag) return false;indegree[now]=-1;//标记前驱数量为 -1for(int i=1; i<=9; i++)//当前节点的后继节点入度 -1if(map[now][i]) indegree[i]--;}return true;
}int main() {char C[32];while(1) {scanf("%s",C);if(!strcmp(C,"ENDOFINPUT")) break;//结束测试数据memset(indegree,0,sizeof(indegree));//数组的初始化memset(map,0,sizeof(map));int screen[16][16];for(int j=0; j<4; j++)for(int k=0; k<4; k++)scanf("%d",&screen[j][k]);scanf("%s",C);for(int j=1; j<=9; j++) {for(int k=0; k<4; k++) {int gx=local[j][0]+dx[k];int gy=local[j][1]+dy[k];int now=screen[gx][gy];//当前屏幕显示的数字if(now!=j&&!map[j][now]) {map[j][now]=1;indegree[now]++;}}
// printf("**********\n");}if(TopoSort()) puts("THESE WINDOWS ARE CLEAN");else puts("THESE WINDOWS ARE BROKEN");}return 0;
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