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  • 【POJ】[2585]Window Pains

    Window Pains
    Time Limit: 1000MS   Memory Limit: 65536K

    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: 
    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
    
    

    理解之后算是道比较水的题。

    因为对于每一个网格,如果网格内的数字不是相同的,则网格内的数字必然出现的比应该出现的那个晚,这样把map遍历一遍,便可得到关于1~9的有向图,剩下就是看其能否topo排序,如果无环则这个图形就能出现。

    #include<stdio.h>
    #include<string.h>
    char s[20];
    int map[5][5];
    int edge[12][12];
    int cnt[12];
    bool topo() {
    	int num=0;
    	for(int i=1; i<=9; i++) {
    		for(int j=1; j<=9; j++) {
    			if(cnt[j]==0) {
    				cnt[j]--;
    				num++;
    				for(int k=1; k<=9; k++) {
    					if(edge[j][k]==1) {
    						cnt[k]--;
    					}
    				}
    				break;
    			}
    		}
    	}
    	return num==9;
    }
    int main() {
    	while(scanf("%s",s),strcmp(s,"ENDOFINPUT")!=0) {
    		memset(edge,0,sizeof(edge));
    		memset(cnt,0,sizeof(cnt));
    		for(int i=1; i<=4; i++)
    			for(int j=1; j<=4; j++)
    				scanf("%d",&map[i][j]);
    
    		scanf("%s",s);
    		for(int i=1; i<=3; i++) {
    			for(int j=1; j<=3; j++)
    				for(int in=0; in<2; in++) {
    					for(int im=0; im<2; im++) {
    
    						if(map[i+in][j+im]==j+(i-1)*3)
    							continue;
    						else if(edge[map[i+in][j+im]][j+(i-1)*3]==0) {
    							edge[map[i+in][j+im]][j+(i-1)*3]=1;
    
    							cnt[j+(i-1)*3]++;
    						}
    					}
    				}
    		}
    		printf("%s
    ",topo()?"THESE WINDOWS ARE CLEAN":"THESE WINDOWS ARE BROKEN");
    	}
    	return 0;
    }

    题目地址:【POJ】[2585]Window Pains

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  • 原文地址:https://www.cnblogs.com/BoilTask/p/12569420.html
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