SDNU 1136.Balloons【山东省第一届ACM】【7月20】

Balloons

原题贴上，不给链接了啊。

Description

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?

Input

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="" 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.

Output

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.

Sample Input

Sample Output

Case 1: 3 2

第一届省赛题。比较简单的DFS。就是找连通器的个数。分别找4方向连通跟8方向连通。代码贴上：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace  std;
int f[110][110],n,m;
int mark1[110][110];
int mark2[110][110];
void DFS1(int x,int y){if(x>-1&&y>-1&&x<n&&y<m&&f[x][y]==1&&mark1[x][y]==0){mark1[x][y]=1;DFS1(x-1,y);DFS1(x,y-1);DFS1(x,y+1);DFS1(x+1,y);}
}
void DFS2(int x,int y){if(x>-1&&y>-1&&x<n&&y<m&&f[x][y]==1&&mark2[x][y]==0){mark2[x][y]=1;DFS2(x-1,y-1);DFS2(x-1,y);DFS2(x-1,y+1);DFS2(x,y-1);DFS2(x,y+1);DFS2(x+1,y-1);DFS2(x+1,y);DFS2(x+1,y+1);}
}
int main()
{int kase=1;while(scanf("%d",&n)==1&&n){memset(mark1,0,sizeof(mark1));memset(mark2,0,sizeof(mark2));for(int i=0;i<n;i++){char s[110];cin>>s;m=strlen(s);for(int j=0;j<m;j++)f[i][j]=s[j]-'0';}int sum1=0,sum2=0;for(int i=0;i<n;i++)for(int j=0;j<m;j++){if(f[i][j]==1&&mark1[i][j]==0){sum1++;DFS1(i,j);}if(f[i][j]==1&&mark2[i][j]==0){sum2++;DFS2(i,j);}}printf("Case %d: %d %d\n\n",kase++,sum1,sum2);}return 0;
}