zoukankan      html  css  js  c++  java
  • Balloons dfs

     
     
     Both Saya and Kudo like balloons. One day, they heard that in the central park, there will be thousands of people fly balloons to pattern a big image.

    They were very interested about this event, and also curious about the image.

    Since there are too many balloons, it is very hard for them to compute anything they need. Can you help them?

    You can assume that the image is an N*N matrix, while each element can be either balloons or blank.

    Suppose element and element B are both balloons. They are connected if:

    i) They are adjacent;
           ii) There is a list of element C1C2, … , Cn, while A and C1 are connected, C1 and C2are connected …Cn and B are connected.

    And a connected block means that every pair of elements in the block is connected, while any element in the block is not connected with any element out of the block.

    To Saya, element A(xa,ya)and B(xb,yb) is adjacent if |xa-xb| + |ya-yb|  1 
           But to Kudo, element A(xa,ya) and element B (xb,yb) is adjacent if |xa-xb|≤1 and |ya-yb|1
           They want to know that there’s how many connected blocks with there own definition of adjacent?

    输入

    The input consists of several test cases.
    The first line of input in each test case contains one integer N (0<N100), which represents the size of the matrix.
    Each of the next N lines contains a string whose length is N, represents the elements of the matrix. The string only consists of 0 and 1, while 0 represents a block and 1represents balloons.
    The last case is followed by a line containing one zero.

    输出

     For each case, print the case number (1, 2 …) and the connected block’s numbers with Saya and Kudo’s definition. Your output format should imitate the sample output. Print a blank line after each test case.

    示例输入

    5
    11001
    00100
    11111
    11010
    10010
    
    0

    示例输出

    Case 1: 3 2
     1 #include<stdio.h>
     2  #include<string.h>
     3  int map[110][110],vis[110][110];
     4  void dfs1(int x,int y)
     5  {
     6      if(map[x][y] == 0 || vis[x][y] == 1)
     7          return;
     8      vis[x][y] = 1;
     9      dfs1(x,y-1);
    10      dfs1(x,y+1);
    11      dfs1(x+1,y);
    12      dfs1(x-1,y);
    13  
    14  }//四连块
    15  void dfs2(int x,int y)
    16  {
    17      if(map[x][y] == 0 || vis[x][y] == 1)
    18          return;
    19      vis[x][y] = 1;
    20      dfs2(x-1,y-1); dfs2(x-1,y);dfs2(x-1,y+1); dfs2(x,y-1);
    21      dfs2(x,y+1); dfs2(x+1,y-1); dfs2(x+1,y);  dfs2(x+1,y+1);
    22  }//八连块
    23  int main()
    24  {
    25  
    26      int n,cnt,sum;
    27      char s[110];
    28      int k =1,i,j;
    29      while(~scanf("%d",&n)&&n)
    30      {
    31          memset(vis,0,sizeof(vis));
    32          for(i = 0; i <= n+1; i++)
    33          {
    34              map[0][i] = 0;
    35              map[i][0] = 0;
    36              map[n+1][i] = 0;
    37              map[i][n+1] = 0;
    38          }外加一道墙
    39          for(i = 1; i <= n;i++)
    40          {
    41              scanf("%s",s);
    42              for(j = 0; j < n; j++)
    43              {
    44                  map[i][j+1] = s[j] - '0';
    45              }
    46          }
    47          cnt = 0,sum = 0;
    48          for(i = 1; i <= n; i++)
    49              for(j = 1; j <= n; j++)
    50                  if(!vis[i][j] && map[i][j])
    51                  {
    52                      cnt++;
    53                      dfs1(i,j);
    54                  }
    55          memset(vis,0,sizeof(vis));
    56          for(i = 1; i <= n; i++)
    57              for(j = 1; j <= n; j++)
    58                  if(!vis[i][j] && map[i][j])
    59                  {
    60                      sum++;
    61                      dfs2(i,j);
    62  
    63                  }
    64         printf("Case %d: %d %d\n\n",k++,cnt,sum);
    65      }
    66      return 0;
    67  
    68  }
    69  
  • 相关阅读:
    .net core 支付宝,微信支付 一
    .net core AES加密解密及RSA 签名验签
    .net core 使用webservice
    .net core 使用X509 私钥加密请求
    .net core mysql entity映射时字符串被截断
    VS 2017 RC .net core ef+ MySql 出现错误
    IdentityServer4 简单使用,包括api访问控制,openid的授权登录,js访问
    iOS面试题之内存管理
    iOS之tableView性能优化/tableView滑动卡顿?
    iOS面试题之runloop
  • 原文地址:https://www.cnblogs.com/LK1994/p/3085566.html
Copyright © 2011-2022 走看看