HDU 1548 a strange lift

A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18422    Accepted Submission(s): 6833

Problem Description

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"?

 

Input

The 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.

 

Output

For 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 53 3 1 2 50

 

Sample Output

3

 思路：依然是典型的最短路径问题，只不过我们把没按一次按钮，不管上升或下降多少层的代价都变成1就好了哦~

<span style="font-size:14px;">import java.util.Scanner;public class Main15482 {public static void main(String[] args){Scanner cin=new Scanner(System.in);int maxnum=100000000;while(cin.hasNext()){int[][] graph=new int[201][201];boolean[] isV=new boolean[201];int N=cin.nextInt();if(N==0)break;int A=cin.nextInt();int B=cin.nextInt();//String[][] path=new String[200][200];for(int i=1;i<=N;i++){                           //初始化邻接矩阵for(int j=1;j<=N;j++) {if(i==j)graph[i][j]=0;else graph[i][j] = maxnum; //path[i][j]="";}}for(int i=1;i<=N;i++){int floNum=cin.nextInt();                //根据读入值，更新矩阵值。folnum代表上下的层数if((i+floNum)<=N)graph[i][i+floNum]=1;if((i-floNum)>=1)graph[i][i-floNum]=1;}/*for(int i=1;i<=N;i++)for(int j=1;j<=N;j++){if(graph[i][j]!=maxnum)graph[i][j]="->"+new Integer(j).toString();else path[i][j]="";}*//*for(int k=1;k<=N;k++)for(int i=1;i<=N;i++)for(int j=1;j<=N;j++)if(graph[i][k]+graph[k][j]<graph[i][j]){graph[i][j]=graph[i][k]+graph[k][j];graph[i][j]=graph[i][k]+graph[k][j];}*/isV[0]=true;int min=0;int v=0;for(int i=0;i<N;i++){min=maxnum;for(int j=1;j<=N;j++){if(isV[j]!=true&&graph[A][j]<min){ min=graph[A][j]; v=j;}}isV[v]=true;for(int j=1;j<=N;j++){if(isV[j]!=true&&graph[A][j]>graph[A][v]+graph[v][j]){graph[A][j]=graph[A][v]+graph[v][j];//path[0][j]=path[0][v]+path[v][j];}}}if(graph[A][B]!=maxnum)System.out.println(graph[A][B]);else System.out.println(-1);}}}</span><span style="color:#cc66cc;font-size:18px; font-weight: bold;"></span>