标签：|| col 不能 his enter NPU https fas continue
Input输入数据的第一行是一个正整数K,表明测试数据的数量.每组测试数据的第一行是四个正整数A,B,C和T(1<=A,B,C<=50,1<=T<=1000),它们分别代表城堡的大小和魔王回来的时间.然后是A块输入数据(先是第0块,然后是第1块,第2块......),每块输入数据有B行,每行有C个正整数,代表迷宫的布局,其中0代表路,1代表墙.(如果对输入描述不清楚,可以参考Sample Input中的迷宫描述,它表示的就是上图中的迷宫)
特别注意:本题的测试数据非常大,请使用scanf输入,我不能保证使用cin能不超时.在本OJ上请使用Visual C++提交.
Output对于每组测试数据,如果Ignatius能够在魔王回来前离开城堡,那么请输出他最少需要多少分钟,否则输出-1.
Sample Input
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
11
对java不友好啊，加个快速输入输出
代码：
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{//快速输入 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);//输入 static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));//输出 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(); } }
标签：|| col 不能 his enter NPU https fas continue
原文地址：https://www.cnblogs.com/qdu-lkc/p/12247965.html
