A strange lift

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 5
3 3 1 2 5
0
Sample Output
3

题目大意：

　　　　从前有一栋楼，楼里面有一座电梯，电梯非常奇葩。

　　　　一栋楼有N层(1-N)，楼层i的电梯可以去i-k[i]和i+k[i]层，(k数组题目给定)(即按了一次按钮)

　　　　求从A层到B层最少的按按钮次数。

解题思路：

　　　　BFS可解，做过类似的题，所以这里使用dijkstra算法。

　　　　最短路径dijkstra算法可解，begin为i，end为i+k[i]&&i-k[i]，边的权值为1。求从A到B的最短路径就是答案。(注意为有向图)

Code:

 #include<stdio.h>

 #include<limits.h>

 #include<iostream>

 #include<string.h>

 #define MAXN 200

 using namespace std;

 int edge[MAXN+][MAXN+];

 int dis[MAXN+];

 bool vis[MAXN+];

 int T,S,D,N,k;

 void dijkstra(int begin)

 {

     memset(vis,,sizeof(vis));

     for (int i=; i<=T; i++)

         dis[i]=INT_MAX;

     dis[begin]=;

     for (int t=; t<=T; t++)

     {

         vis[begin]=;

         for (int i=; i<=T; i++)

             if (!vis[i]&&edge[begin][i]!=INT_MAX&&dis[begin]+edge[begin][i]<dis[i])

                 dis[i]=dis[begin]+edge[begin][i];

         int min=INT_MAX;

         for (int j=; j<=T; j++)

             if (!vis[j]&&min>dis[j])

             {

                 min=dis[j];

                 begin=j;

             }

     }

 }

 int main()

 {

     int begin,end;

     while (cin>>T)

     {

         if (T==) break;

         for (int i=;i<=MAXN;i++)

             for (int j=;j<=MAXN;j++)

                 edge[i][j]=INT_MAX;

         scanf("%d %d",&begin,&end);

         int t;

         for (int i=;i<=T;i++)

         {

             scanf("%d",&t); //注意是有向图！一开始因为这个WA了好几次。

             if (i+t<=T) edge[i][i+t]=;

             if (i-t>=) edge[i][i-t]=;

         }

         dijkstra(begin);

         if (dis[end]!=INT_MAX) printf("%d\n",dis[end]);

         else printf("-1\n");

     }

     return ;

 }