Description

Input

輸入包含不超過10000 組資料。每組資料包含6個整數r1, c1, r2, c2, r3, c3 (1<=r1, c1, r2, c2, r3, c3<=8). 三個格子A, B, C保證各不相同。

Output

對於每組資料，輸出測試點編號和最少步數。

Sample Input

1 1 8 7 5 6
1 1 3 3 2 2

Sample Output

Case 1: 7
Case 2: 3

程式碼如下：

#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;
const int INF = 1000000;
typedef pair<int,int> P;
int ans;
int dir[8][2] = {{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};
int d[9][9];
int sx,sy,ex,ey,cx,cy;
int bfs()
{
	queue <P> que;
	for(int i = 1;i < 9;i++){
		for(int j = 1;j < 9;j++){
			d[i][j] = INF;
		}
	}
	que.push(P(sx,sy));
	d[sx][sy] = 0;
	while(que.size()){
		P p = que.front();
		que.pop();
		if(p.first == ex && p.second == ey)
			break;
		for(int i = 0;i < 8;i++){
			int nx = p.first + dir[i][0];
			int ny = p.second + dir[i][1];
			if(nx < 1 || 9 <= nx || ny < 1 || 9 <= ny || (nx == cx && ny == cy) || d[nx][ny] != INF){
				continue;
			}else{
				que.push(P(nx,ny));
				d[nx][ny] = d[p.first][p.second] + 1;
			}
		}
	}
	return d[ex][ey];
}

int main()
{
	int ncase,i,j,k;
	while(scanf("%d %d %d %d %d %d",&sx,&sy,&ex,&ey,&cx,&cy) != EOF){
		ans = bfs();
		printf("Case %d: %d\n",++ncase,ans);
	} 
	return 0;
}