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ZOJ 3864 Quiz for EXO-L

时间:2015-07-10 09:30:10      阅读:113      评论:0      收藏:0      [点我收藏+]

标签:思维   zoj   图形题   搜索   

题目链接:Quiz for EXO-L


题面:

B - Quiz for EXO-L
Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Description

Exo (Korean: ??; Chinese:爱咳嗽; often stylized as EXO) is a Chinese-South Korean boy band based in Seoul, South Korea. Formed by SM Entertainment in 2011, the group consists of twelve members separated into two subgroups, EXO-K and EXO-M, performing music in Korean and Mandarin, respectively. The group officially debuted on April 8, 2012, with Suho, Baekhyun, Chanyeol, D.O, Kai, and Sehun under the Korean subgroup while Xiumin, Luhan, Kris, Lay, Chen, and Tao are under the Mandarin group. Kris filed a lawsuit against S.M. to be removed from the group, and Luhan followed suit in October 2014. The band now promotes with 10 members since mid-October 2014, which consists of 8 Korean and 2 Chinese members.

In their first album, each member of EXO has a kind of super power. For example, Tao can control time and Lay has a super power of healing. The badges are the symbols of their super powers. Each member of EXO has an unique badge. The following table shows the shapes of their badges.

技术分享 技术分享 技术分享 技术分享
Baekhyun Chanyeol Chen D.O
技术分享 技术分享 技术分享 技术分享
Kai Kris Lay Luhan
技术分享 技术分享 技术分享 技术分享
Sehun Suho Tao Xiumin

EXO’s official fanclub name has been announced to be EXO-L. EXO-L is short for EXO-LOVE, L being the letter between M and K in the alphabet, signifying all the fans‘ love for both EXO-K and EXO-M of EXO, and also holds the meaning of ‘EXO and the fans are one‘ like their team slogan ‘WE ARE ONE.‘

You should help an EXO-L pass a quiz about image recognition. The quiz requires a program to recognize the badges. The input is one of the above 12 images with rotation, and the output should be the owner of the badge in the image. The badge is rotated at an unknown angle. And each image is a square matrix. The badges will never touch or exceed the boundary of the image.

Since the images are large, we will compress the images. Each pixel in the image is either 1(white) or 0(black) after binarization. The compress method is simple:

  • Concatenate the pixels in the row first order to a string.
  • Split the string to some runs with same characters, and two adjacent runs have different characters.
  • Output the length of each run.

For example, an image in size 6×6:

111111
100101
100011
110111
101001
111111
We can compress it as the following: 
  • Concatenate the pixels in the row first order, then we can get
    111111100101100011110111101001111111
    
  • The runs are
    1111111 00 1 0 11 000 1111 0 1111 0 1 00 1111111
    
  • So the result of compressed images is
    7 2 1 1 2 3 4 1 4 1 1 2 7
    
Note that in this task the first run and the last run is always in white.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains 2 integers n and mn is the size of the square image. m is the number of the runs in the compressed images. (100≤n ≤ 900)

The second line contains m positive integers, indicating the length of the each runs in the compressed image. (The sum of the integers isn2).

Output

the length of each run.

For example, an image in size 6×6:

111111
100101
100011
110111
101001
111111
We can compress it as the following: 
  • Concatenate the pixels in the row first order, then we can get
    111111100101100011110111101001111111
    
  • The runs are
    1111111 00 1 0 11 000 1111 0 1111 0 1 00 1111111
    
  • So the result of compressed images is
    7 2 1 1 2 3 4 1 4 1 1 2 7
    
Note that in this task the first run and the last run is always in white.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains 2 integers n and mn is the size of the square image. m is the number of the runs in the compressed images. (100≤n ≤ 900)

The second line contains m positive integers, indicating the length of the each runs in the compressed image. (The sum of the integers isn2).

Output

For each case, output the owner of the badge in one line. Your output should be formated as one of following words: "Suho", "Baekhyun", "Chanyeol", "D.O", "Kai", "Sehun", "Xiumin", "Luhan", "Kris", "Lay", "Chen", and "Tao".

Sample Input

2
100 443
1445 1 99 2 97 3 97 4 6 1 88 5 5 2 88 5 4 4 86 6 4 4 73 1 12 6 3 6 72 4 8 6 5 5 72 6 6 6 5 6 72 7 3 6 6 6 12 1 59 16 7 6 9 3 60 14 8 6 6 6 61 13 9 6 3 7 65 12 7 16 8 1 45 3 10 12 5 15 6 4 45 5 9 14 3 12 6 6 45 6 8 15 2 9 9 5 47 5 7 17 1 7 10 6 47 6 6 17 2 3 13 5 48 6 5 18 17 6 49 5 5 8 1 10 5 3 8 5 50 6 3 8 5 7 3 7 6 5 50 6 3 8 7 18 2 5 52 5 3 7 10 23 52 6 2 7 13 20 52 6 3 6 15 17 53 7 2 5 19 15 49 10 2 5 21 16 44 12 2 4 24 17 38 12 6 3 25 18 34 12 6 4 26 7 1 13 31 9 7 6 25 7 5 10 34 4 6 9 25 7 8 5 46 8 25 8 58 9 25 7 59 8 25 8 59 8 25 7 48 4 7 8 26 7 8 2 35 9 5 8 25 6 7 7 33 11 3 8 25 4 6 11 35 18 25 3 5 15 36 16 25 3 3 15 40 17 21 4 2 14 44 17 19 5 2 10 49 17 16 6 2 8 51 20 13 6 2 6 52 23 11 7 2 4 53 5 2 18 8 8 2 5 51 6 5 8 2 7 6 7 3 5 51 5 8 4 5 9 2 8 4 4 50 6 17 18 5 5 49 5 13 3 2 18 5 5 48 6 11 5 3 16 7 5 47 6 9 8 2 16 7 5 47 5 7 11 3 14 9 4 46 3 8 14 4 13 9 4 45 1 8 16 6 14 8 2 54 16 9 13 62 9 3 5 8 15 60 6 6 5 8 16 58 5 9 5 7 5 2 9 58 2 12 5 6 6 5 7 72 5 5 5 8 5 72 5 4 6 11 2 73 3 5 5 87 3 5 5 87 3 5 4 89 1 6 4 97 3 97 2 98 2 1445
100 451
1449 2 96 4 96 5 94 6 93 6 94 5 93 6 93 6 10 4 78 7 7 13 72 6 6 18 80 21 62 2 14 24 58 3 14 26 55 3 13 15 6 8 54 4 10 20 8 5 51 5 8 25 8 3 51 4 7 29 7 3 49 5 6 31 7 2 48 5 6 15 4 14 7 1 48 5 5 15 9 11 54 6 4 15 12 10 53 5 4 16 14 8 53 5 3 16 16 8 51 6 3 8 1 6 18 8 50 6 2 8 2 6 19 7 50 6 2 7 3 5 7 21 5 1 43 6 1 7 4 5 4 24 4 2 43 6 1 7 4 5 2 26 4 3 43 12 4 5 2 28 3 3 43 12 4 5 16 14 3 4 42 11 5 5 19 12 3 4 42 10 5 5 15 1 5 10 3 5 41 10 6 3 16 1 7 8 4 5 40 9 7 3 16 2 7 8 3 7 32 1 6 8 7 3 16 2 8 7 4 7 29 5 4 8 7 3 16 3 7 8 4 6 29 6 4 7 7 3 16 3 8 7 5 5 29 7 3 8 7 2 16 3 8 8 5 3 31 6 4 8 6 2 16 4 7 8 40 6 4 8 6 1 16 4 6 10 40 6 3 9 6 1 15 4 6 10 42 4 3 11 20 4 6 10 42 4 4 12 18 4 6 11 42 3 4 15 15 4 5 12 43 2 4 26 3 5 5 12 43 2 4 25 4 5 4 6 1 6 44 1 5 6 2 14 6 5 4 6 1 6 44 1 5 7 5 6 9 6 3 7 1 6 50 8 19 6 2 7 2 6 51 8 18 6 1 7 3 6 51 9 16 15 3 6 52 10 13 15 4 5 54 11 11 14 5 5 47 1 6 14 6 15 6 4 49 1 6 33 6 5 49 2 6 31 7 4 50 3 7 27 8 4 52 4 6 25 9 3 53 6 7 20 10 3 55 29 12 2 57 26 14 1 60 23 78 20 6 5 71 15 7 8 72 10 8 7 92 7 92 6 93 6 94 6 94 5 95 4 97 2 1448

Sample Output

Xiumin
Sehun

Hint

The images in the sample input are visualized as:

技术分享 技术分享
Sample 1
Sample 2


解题:

这绝对是我目前为止做过的最掉炸天的图形处理题了。之前毫无思路,看了别人的题解之后,才知道去判断图中黑白联通块的数量,可以判断出十张图,然而有两张图是一样的,即给出的两组样例。只能说题目设计的很有心机,通过给定的两组样例,数黑白块数量,拿白块除以黑块,一个值为4.3,一个值为4.8,以4.5为界判断即可。(深搜会爆栈,然而我应该想到,所以用广搜)。


代码:(代码写得很挫,看看就好..)

#include <iostream>
#include <cstdio>
#include <fstream>
#include <cstring>
#include <queue>
using namespace std;
bool status1[1000][1000],status2[1000][1000];
bool vis1[1000][1000],vis2[1000][1000];
int t,n,m,tmp,cnt_black,cnt_white,flag,row,column,whiteblock,blackblock,ans;
void bfs1(int x,int y)
{
	int xx,yy;
	queue <int> q1;
	queue <int> q2;
	status1[x][y]=0;
    q1.push(x);q2.push(y-1);
	q1.push(x);q2.push(y+1);
	q1.push(x-1);q2.push(y-1);
	q1.push(x-1);q2.push(y+1);
	q1.push(x-1);q2.push(y);
    q1.push(x+1);q2.push(y-1);
	q1.push(x+1);q2.push(y+1);
	q1.push(x+1);q2.push(y);
	while(!q1.empty())
	{
      xx=q1.front();
      q1.pop();
	  yy=q2.front();
	  q2.pop();
	  if(xx<0||yy<0||xx>901||yy>901)
	    continue;
	  if(status1[xx][yy]==0)continue;
	  status1[xx][yy]=0;
      q1.push(xx);q2.push(yy-1);
	  q1.push(xx);q2.push(yy+1);
	  q1.push(xx-1);q2.push(yy-1);
   	  q1.push(xx-1);q2.push(yy+1);
	  q1.push(xx-1);q2.push(yy);
      q1.push(xx+1);q2.push(yy-1);
	  q1.push(xx+1);q2.push(yy+1);
	  q1.push(xx+1);q2.push(yy);
	}
}

void search1()
{
   for(int i=1;i<=n;i++)
   {
	   for(int j=1;j<=n;j++)
	   {
		   if(status1[i][j]==1)
		   {
			   bfs1(i,j);
			   whiteblock++;
		   }
	   }
   }
}
void bfs2(int x,int y)
{
	int xx,yy;
	queue <int> q1;
	queue <int> q2;
	status2[x][y]=1;
    q1.push(x);q2.push(y-1);
	q1.push(x);q2.push(y+1);
	q1.push(x-1);q2.push(y-1);
	q1.push(x-1);q2.push(y+1);
	q1.push(x-1);q2.push(y);
    q1.push(x+1);q2.push(y-1);
	q1.push(x+1);q2.push(y+1);
	q1.push(x+1);q2.push(y);
	while(!q1.empty())
	{
      xx=q1.front();
	  q1.pop();
	  yy=q2.front();
	  q2.pop();
	  if(xx<0||yy<0||xx>901||yy>901)
	    continue;
	  if(status2[xx][yy]==1)continue;
	  status2[xx][yy]=1;
      q1.push(xx);q2.push(yy-1);
	  q1.push(xx);q2.push(yy+1);
	  q1.push(xx-1);q2.push(yy-1);
   	  q1.push(xx-1);q2.push(yy+1);
	  q1.push(xx-1);q2.push(yy);
      q1.push(xx+1);q2.push(yy-1);
	  q1.push(xx+1);q2.push(yy+1);
	  q1.push(xx+1);q2.push(yy);
	}
}

void search2()
{
   for(int i=1;i<=n;i++)
   {
	   for(int j=1;j<=n;j++)
	   {
		   if(status2[i][j]==0)
		   {
			   bfs2(i,j);
			   blackblock++;
		   }
	   }
   }
}

int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
        whiteblock=blackblock=cnt_black=cnt_white=0;
		flag=row=column=1;
		memset(status1,0,sizeof(status1));
		memset(status2,0,sizeof(status2));
		memset(vis1,-1,sizeof(vis1));
		memset(vis2,-1,sizeof(vis2));
		for(int i=0;i<m;i++)
		{
			scanf("%d",&tmp);
			if(flag==1)
		   {
			   for(int j=0;j<tmp;j++)
			  {
                 status1[row][column]=status2[row][column]=1;
			     cnt_white++;
			     if(column==n)
			     {
				   row++;
				   column=0;
			     }
				 column++;
			  }
			   flag*=-1;
		   }
			else
			{
               for(int j=0;j<tmp;j++)
			   {
			     cnt_black++;
			     if(column==n)
			     {
				     row++;
				     column=0;
			     }
				 column++;
		   	   }
               flag*=-1;
			}
		}
		search1();
		search2();
		ans=blackblock*100+whiteblock;
		if(ans==902)printf("Baekhyun\n");
		else if(ans==501)printf("Chanyeol\n");
		else if(ans==103)printf("Chen\n");
		else if(ans==102)printf("D.O\n");
		else if(ans==213)printf("Kai\n");
        else if(ans==301)printf("Kris\n");
		else if(ans==602)printf("Lay\n");
        else if(ans==508)printf("Luhan\n");
		else if(ans==208)printf("Suho\n");
		else if(ans==204)printf("Tao\n");
		else
		{
           if(2*cnt_white>(cnt_black*9))
			   printf("Xiumin\n");
		   else printf("Sehun\n");
		}

	}
}



版权声明:本文为博主原创文章,未经博主允许不得转载。

ZOJ 3864 Quiz for EXO-L

标签:思维   zoj   图形题   搜索   

原文地址:http://blog.csdn.net/david_jett/article/details/46822889

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