hdu1548

A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 25624    Accepted Submission(s): 9211

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

题目大意：现在有一个电梯，每一层楼有一个数k，代表可以从这楼上去k层或者下去k层，现在给出这栋楼共有多少层和开始所在楼层目的楼层，求从a层到b层需要坐多少次电梯。

题目分析：直接BFS暴力解决，每次压入可能的下一种情况（1=<楼层数<=n），如果下一种情况楼层超出范围则不必加入。

代码：

#include <iostream>#include <string.h>#include <queue>using namespace std;int main(){    int n,start,stop,now,next;    int visit[300];    int k[300];    queue<int>que;    while(cin>>n&&n)    {        cin>>start>>stop;        for(int i=1;i<=n;i++){            cin>>k[i];        }        memset(visit,0,sizeof(visit));        while(!que.empty())        {            que.pop();        }        now=start;        visit[now]=1;        que.push(now);        while(!que.empty())        {            now=que.front();            que.pop();            if(now==stop){                cout<<visit[stop]-1<<endl;                break;            }            next=now+k[now];            if(!visit[next]&&next>=1&&next<=n){                que.push(next);                visit[next]=visit[now]+1;            }            next=now-k[now];            if(!visit[next]&&next>=1&&next<=n){                que.push(next);                visit[next]=visit[now]+1;            }            if(que.empty()){                cout<<-1<<endl;            }        }    }    return 0;}