HDU1548-A strange lift

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本文探讨了一个具有特殊楼层跳跃功能的电梯问题，通过使用广度优先搜索算法（BFS），寻找从任意起点到达终点所需的最少按钮按压次数。问题设定在一个包含N层的大楼中，每层楼都有一个特定的跳跃数Ki。

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A strange lift

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21686 Accepted Submission(s): 7945

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

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>

using namespace std;

int n,p,q;
int visit[205],a[205];
struct node
{
    int x,step;
}s;

int bfs()
{
    queue<node>Q;
    node pre,next;
    s.x=p;s.step=0;
    Q.push(s);
    while(!Q.empty())
    {
        pre=Q.front();
        Q.pop();
        if(pre.x==q) return pre.step;
        next.step=pre.step+1;
        next.x=pre.x+a[pre.x];
        if(next.x<=n&&!visit[next.x]) visit[next.x]=1,Q.push(next);
        next.x=pre.x-a[pre.x];
        if(next.x>=1&&!visit[next.x]) visit[next.x]=1,Q.push(next);
    }
    return -1;
}

int main()
{
    while(~scanf("%d",&n)&&n)
    {
        memset(visit,0,sizeof visit);
        scanf("%d %d",&p,&q);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        visit[p]=1;
        printf("%d\n",bfs());
    }
    return 0;
}