There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will go up Ki floor,i.e,you will go to the i+Ki th floor,as the same, if you press the button "DOWN" , you will go down Ki floor,i.e,you will go to the i-Ki th floor. Of course, the lift can't go up high than N,and can't go down lower than 1. For example, there is a buliding with 5 floors, and k1 = 3, k2 = 3,k3 = 1,k4 = 2, k5 = 5.Begining from the 1 st floor,you can press the button "UP", and you'll go up to the 4 th floor,and if you press the button "DOWN", the lift can't do it, because it can't go down to the -2 th floor,as you know ,the -2 th floor isn't exist.
Here comes the problem: when you are on floor A,and you want to go to floor B,how many times at least he has to press the button "UP" or "DOWN"?
InputThe input consists of several test cases.,Each test case contains two lines.
The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,The second line consist N integers k1,k2,....kn.
A single 0 indicate the end of the input.OutputFor each case of the input output a interger, the least times you have to press the button when you on floor A,and you want to go to floor B.If you can't reach floor B,printf "-1".Sample Input
5 1 5 3 3 1 2 5 0
Sample Output
3
用vis[]陣列記錄已經走過哪層
廣搜注意事項：對于有多個輸入
1.vis陣列一定要初始化
2.佇列一定要清空
3.初始點一定要標記vis=true
4.分清楚到底幾個方向，是4個方向，還是6個方向，for（int i=0;i<4||i<6;i++）
5.遍歷完每種情況一定要入隊
代碼：
import java.util.*; class Node{ int x; long times; public Node(int x,long times){ this.x=x; this.times=times; } } public class Main{ static ArrayDeque<Node> q=new ArrayDeque<Node>(); static final int N=205; static int val[]=new int[N]; static boolean vis[]=new boolean[N]; static int n,a,b; static long bfs(){ while(!q.isEmpty()) q.poll(); Arrays.fill(vis, false); vis[a]=true;; q.offer(new Node(a,0)); while(!q.isEmpty()){ Node t=q.poll(); if(t.x==b) return t.times; for(int i=-1;i<=1;i+=2){ int xx=t.x+i*val[t.x]; if(xx<=0||xx>n ||vis[xx]) continue; vis[xx]=true; q.offer(new Node(xx,t.times+1)); } } return -1; } public static void main(String[] args) { Scanner scan=new Scanner(System.in); while(scan.hasNext()){ n=scan.nextInt(); if(n==0) break; a=scan.nextInt(); b=scan.nextInt(); for(int i=1;i<=n;i++) val[i]=scan.nextInt(); System.out.println(bfs()); } } }
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