[2010山东ACM省赛] Balloons（搜索）

Balloons

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

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,
… , Cn,
while A and C1 are
connected, C1 and C2 are
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<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

解题思路：
有一段时间没有做搜索题了，这一题花了不少时间，汗。这题属于简单题，两个搜索，一个搜索是四个方向，另一个搜索是八个方向，对应相同的地图，搜索了几次，就有多少个块。

代码：

#include <iostream>
#include <string.h>
using namespace std;

char maze[102][102];
char maze2[102][102];
int n;

int dx[4]={0,0,-1,1};//四个方向
int dy[4]={-1,1,0,0};

int dxx[8]={0,0,-1,1,-1,1,-1,1};//八个方向
int dyy[8]={1,-1,0,0,1,1,-1,-1};

void dfs(int x,int y)
{
if(x<1||x>n||y<1||y>n)
return ;
if(maze[x][y]=='0')
return;
maze[x][y]='0';
for(int i=0;i<4;i++)
{
int a=x+dx[i];//不能写成 x=x+dx[i]
int b=y+dy[i];
dfs(a,b);
}
}
void dfs2(int x,int y)
{
if(x<1||x>n||y<1||y>n)
return ;
if(maze2[x][y]=='0')
return;
maze2[x][y]='0';
for(int i=0;i<8;i++)
{
int a=x+dxx[i];
int b=y+dyy[i];
dfs2(a,b);
}
}
int main()
{

string s;
int c=1;
while(cin>>n&&n)
{
for(int i=1;i<=n;i++)
{
cin>>s;
for(int j=0;j<s.length();j++)
{
maze[i][j+1]=s[j];
maze2[i][j+1]=s[j];
}
}
int count1=0;
int count2=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
if(maze[i][j]=='1')
{
dfs(i,j);
count1++;
}
if(maze2[i][j]=='1')
{
dfs2(i,j);
count2++;
}
}
cout<<"Case "<<c++<<": "<<count1<<" "<<count2<<endl;
cout<<endl;
}
return 0;
}