Input輸入數據的第一行是一個正整數K,代表測試數據的數量.每組測試數據的第一行是四個正整數A,B,C和T(1<=A,B,C<=50,1<=T<=1000),它們分別表明城堡的大小和魔王回來的時間.而後是A塊輸入數據(先是第0塊,而後是第1塊,第2塊......),每塊輸入數據有B行,每行有C個正整數,表明迷宮的佈局,其中0表明路,1表明牆.(若是對輸入描述不清楚,能夠參考Sample Input中的迷宮描述,它表示的就是上圖中的迷宮)
特別注意:本題的測試數據很是大,請使用scanf輸入,我不能保證使用cin能不超時.在本OJ上請使用Visual C++提交.
Output對於每組測試數據,若是Ignatius可以在魔王回來前離開城堡,那麼請輸出他最少須要多少分鐘,不然輸出-1.
Sample Inputjava
1 3 3 4 20 0 1 1 1 0 0 1 1 0 1 1 1 1 1 1 1 1 0 0 1 0 1 1 1 0 0 0 0 0 1 1 0 0 1 1 0
Sample Output佈局
11
對java不友好啊,加個快速輸入輸出
代碼:
import java.util.*; import java.util.Comparator; import java.io.BufferedReader; import java.io.BufferedWriter; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.util.StringTokenizer; class Node{ int x; int y; int z; int step; public Node(int x,int y,int z,int step){ this.x=x; this.y=y; this.z=z; this.step=step; } } public class Main{ static class FastScanner{//快速輸入 BufferedReader br; StringTokenizer st; public FastScanner(InputStream in) { br = new BufferedReader(new InputStreamReader(in),16384); eat(""); } public void eat(String s) { st = new StringTokenizer(s); } public String nextLine() { try { return br.readLine(); } catch (IOException e) { return null; } } public boolean hasNext() { while(!st.hasMoreTokens()) { String s = nextLine(); if(s==null) return false; eat(s); } return true; } public String next() { hasNext(); return st.nextToken(); } public int nextInt() { return Integer.parseInt(next()); } } static FastScanner scan = new FastScanner(System.in);//輸入 static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));//輸出 static final int N=55; static int map[][][]=new int[N][N][N]; static int a,b,c,T; static boolean vis[][][]=new boolean[N][N][N]; static ArrayDeque<Node> q=new ArrayDeque<Node>(); static int dx[]={1,-1,0,0,0,0}; static int dy[]={0,0,1,-1,0,0}; static int dz[]={0,0,0,0,1,-1}; static int bfs(){ while(!q.isEmpty()) q.poll(); vis[0][0][0]=true; q.offer(new Node(0,0,0,0)); while(!q.isEmpty()){ Node s=q.poll(); if(s.x==a-1 && s.y==b-1 &&s.z==c-1 &&s.step>T) return -1; if(s.x==a-1 && s.y==b-1 &&s.z==c-1 &&s.step<=T) return s.step; for(int i=0;i<6;i++){ int xx=s.x+dx[i]; int yy=s.y+dy[i]; int zz=s.z+dz[i]; if(xx<0||yy<0||zz<0||xx>=a||yy>=b||zz>=c) continue; if(vis[xx][yy][zz]||map[xx][yy][zz]==1) continue; vis[xx][yy][zz]=true; q.offer(new Node(xx,yy,zz,s.step+1)); } } return -1; } public static void main(String[] args) { // Scanner scan=new Scanner(System.in); int t=scan.nextInt(); while(t-->0){ a=scan.nextInt(); b=scan.nextInt(); c=scan.nextInt(); T=scan.nextInt(); for(int i=0;i<a;i++) for(int j=0;j<b;j++) for(int k=0;k<c;k++){ map[i][j][k]=scan.nextInt(); vis[i][j][k]=false; } out.println(bfs()); } out.flush(); } }