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"?

 

 

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;
}