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 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.
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
HINT
Source
题目链接：http://acm.upc.edu.cn/problem.php?id=1929
这道题不是很难，可能初次做的同学会觉得无从下手，思路很简单
第一步：输入
第二部：对输入的数据加墙，即把输入的数据用0括起来；假设输入
1 1 0 0 1
0 0 1 0 0
1 1 1 1 1
1 1 0 1 0
1 0 0 1 0
加墙：
0 0 0 0 0 0 0
0 1 1 0 0 1 0
0 0 0 1 0 0 0
0 1 1 1 1 1 0
0 1 1 0 1 0 0
0 1 0 0 1 0 0
0 0 0 0 0 0 0
第三步：定义两个标志位数组a[n][n]，b[n][n],分别对应第一种搜索方式，和第二种搜索方式
第四步：对于第一种搜索方式：上下左右四个方向搜索，跳出条件是“无路可走”或者“标志位为1”（代表已经搜索过）——if(a[x][y]==0||b[x][y]==1){return ;}
对于第二种搜索方式：八个方向搜索，就跳出条件与第一种一致
代码：
import java.util.Scanner;
public class Main {
public static int a[][];
public static int b[][];
public static int c[][];
public static void dfs1(int x,int y){
if(a[x][y]==0||b[x][y]==1){
return ;
}
b[x][y]=1;
dfs1(x,y+1);
dfs1(x,y-1);
dfs1(x-1,y);
dfs1(x+1,y); }
public static void dfs2(int x,int y){
if(a[x][y]==0||c[x][y]==1){
return ;
}
c[x][y]=1;
dfs2(x,y+1);
dfs2(x,y-1);
dfs2(x-1,y);
dfs2(x+1,y);
dfs2(x+1,y+1);
dfs2(x-1,y-1);
dfs2(x-1,y+1);
dfs2(x+1,y-1);
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int d=1;
while(sc.hasNextInt()){ int t=sc.nextInt();
if(t==0)
break;
a=new int[t+2][t+2];
b=new int[t+2][t+2];
c=new int[t+2][t+2];
for(int i=1;i<t+1;i++){
String s=sc.next();
for(int j=1;j<t+1;j++){
a[i][j]=s.charAt(j-1)-'0';
}
}
for(int i=0;i<t+2;i++){
for(int j=0;j<t+2;j++){
System.out.print(a[i][j]+" ");
}
System.out.println();
}
int con1=0,con2=0;
for(int i=1;i<t+1;i++){
for(int j=1;j<t+1;j++){
if(a[i][j]!=0&&b[i][j]==0){
con1++;
dfs1(i,j);
}
}
}
for(int i=1;i<t+1;i++){
for(int j=1;j<t+1;j++){
if(a[i][j]!=0&&c[i][j]==0){
con2++;
dfs2(i,j);
}
}
}
System.out.println("Case "+d+": "+con1+" "+con2);
System.out.println();
d++;
}
}
}