Balloons dfs

最新推荐文章于 2018-12-05 14:15:27 发布

转载最新推荐文章于 2018-12-05 14:15:27 发布 · 64 阅读

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原文链接：http://www.cnblogs.com/LK1994/archive/2013/05/18/3085566.html

本文介绍了一种用于识别由大量气球组成的图案的算法。该算法通过两种不同的连接定义来计算气球形成的连通区域数量，一种是考虑八方向连接，另一种是更宽松的四方向连接标准。通过对矩阵进行深度优先搜索，可以有效地找到并计数这些连通区域。

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

i) They are adjacent;
ii) There is a list of element C₁, C₂, … , C_n, while A and C₁ are connected, C₁ and C₂are connected …C_n 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.

输入

The input consists of several test cases.
The first line of input in each test case contains one integer N (0<N≤100), 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

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

转载于:https://www.cnblogs.com/LK1994/archive/2013/05/18/3085566.html