A strange lift

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"?

[align=left]Input[/align]
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.

[align=left]Output[/align]
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".

[align=left]Sample Input[/align]

5 1 5
3 3 1 2 5
0

[align=left]Sample Output[/align]

3

题意:一个楼层有N层，通过电梯从A层到B层，电梯每层（i）只能选择向上和向下，并且为固定的k[i]（向上或向下），问问至少多少次能够从A层到B层，如果不能到达，输出-1

思路：求最优解法，选择BFS

代码:

#include<queue>
#include<string.h>
#include<iostream>
using namespace std;

const int num = 255;
int floor[num][2];//表示上下楼
bool vis[num];//标识楼层是否到达过
//定义楼层
struct node{
　　int pos;//楼层
　　int cen;//积累到达该层移动的次数
};
node temp;
queue<node> q;
int N,A,B;

int BFS(){
　　memset(vis,false,sizeof(vis));
　　while(!q.empty()){
　　　　temp = q.front();
　　　　q.pop();
　　　　vis[temp.pos] = true;
　　　　if(temp.pos == B){
　　　　　　return temp.cen;
　　　　}
　　　　int up = floor[temp.pos][0];
　　　　int down = floor[temp.pos][1];
　　　　node temp2;
　　　　if(up != -1 && !vis[up]){
　　　　　　temp2.pos = up;
　　　　　　temp2.cen = temp.cen + 1;
　　　　　　q.push(temp2);
　　　　}
　　　　if(down != -1 && !vis[down]){
　　　　　　temp2.pos = down;
　　　　　　temp2.cen = temp.cen + 1;
　　　　　　q.push(temp2);
　　　　}
　　}
　　return -1;
}
int main(){
　　while(cin >> N && N){
　　　　cin >> A >> B;
　　　　memset(floor,-1,sizeof(floor));
　　　　//清空队列
　　　　while(!q.empty())
　　　　　　q.pop();
　　　　for(int i = 1;i <= N;i ++){
　　　　　　int t = 0;
　　　　　　cin >> t;
　　　　　　temp.pos = i;
　　　　　　if(i + t <= N){
　　　　　　　　//上升后到达的层数
　　　　　　　　floor[i][0] = t + i;
　　　　　　}
　　　　　　if(i - t >= 1){
　　　　　　//下降后到达的层数
　　　　　　　　floor[i][1] = i - t;
　　　　　　}
　　　　　　if(i == A){
　　　　　　　　temp.cen = 0;
　　　　　　　　q.push(temp);
　　　　　　}
　　　　}
　　　　int num = BFS();
　　　　cout << num << endl;
　　}

　　return 0;
}