sdut 2152:Balloons（第一届山东省省赛原题，DFS搜索）

Balloons

Time Limit: 1000MS Memory limit: 65536K

题目描述

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 C₂ are 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.
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<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

5
11001
00100
11111
11010
10010

0

示例输出

Case 1: 3 2

提示

 

来源

 2010年山东省第一届ACM大学生程序设计竞赛

 

---

　　DFS搜索，求连通分量的个数。需要注意的是第一个只能是4个方向，而第二个可以有8个方向，即可以斜着走。

　　代码：

 #include <iostream>
 #include <stdio.h>
 using namespace std;
 int n;
 int dx[] = {,-,,};
 int dy[] = {-,,,};
 int Dx[] = {,-,-,-,,,,};
 int Dy[] = {-,-,,,,,,-};

 bool judge(char a[][],int x,int y)
 {
     if(x< || y< || x>=n || y>=n)
         return ;
     if(a[x][y]!='')
         return ;
     return ;
 }
 void dfs1(char a[][],int x,int y)
 {
     a[x][y] = '';
     for(int i=;i<;i++){
         int cx = x+dx[i];
         int cy = y+dy[i];
         if(judge(a,cx,cy))
             continue;
         dfs1(a,cx,cy);
     }
 }
 void dfs2(char a[][],int x,int y)
 {
     a[x][y] = '';
     for(int i=;i<;i++){
         int cx = x+Dx[i];
         int cy = y+Dy[i];
         if(judge(a,cx,cy))
             continue;
         dfs2(a,cx,cy);
     }
 }
 int main()
 {
     int Count=;
     while(cin>>n){
         if(n==) break;
         int num1 = ,num2 = ;
         char a[][];
         char b[][];
         for(int i=;i<n;i++){
             cin>>a[i];
         }
         for(int i=;i<n;i++)
             for(int j=;j<n;j++)
                 b[i][j]=a[i][j];
         int i,j;
         for(i=;i<n;i++)
             for(j=;j<n;j++){
                 if(a[i][j]==''){
                     num1++;
                     dfs1(a,i,j);
                 }
             }
         for(i=;i<n;i++)
             for(j=;j<n;j++){
                 if(b[i][j]==''){
                     num2++;
                     dfs2(b,i,j);
                 }
             }
         cout<<"Case "<<Count++<<": "<<num1<<' '<<num2<<endl<<endl;
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013