HDU 1548 A strange lift

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

BFS

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：

坐电梯：每一层都可以上或者下k层，问你能不能从第a层到第b层，如果能，最少要坐几次电梯？如果不能，输出-1；

解题思路：

bfs把上楼和下楼都走一遍，如果能到，就输出最先到达的步数，如果不能到，输出-1；

完整代码：

#include<bits/stdc++.h>
using namespace std;
int vis[210];
int a[210];
int n,ans;
struct node
{
    int x;
    int num;
};
int bfs(int st,int en)
{
    queue<node>q;
    node now,next;
    now.x=st;
    ans=-1;
    now.num=0;
    q.push(now);
    vis[now.x]=1;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        if(now.x==en)
        {
            ans=now.num;
            break;
        }
        next.x=now.x+a[now.x];
        if(next.x>=1&&next.x<=n&&vis[next.x]==0)
        {
            vis[next.x]=1;
            next.num=now.num+1;
            q.push(next);
        }
        next.x=now.x-a[now.x];
        if(next.x>=1&&next.x<=n&&vis[next.x]==0)
        {
            vis[next.x]=1;
            next.num=now.num+1;
            q.push(next);
        }
    }
    return ans;
}
int main()
{
    while(scanf("%d",&n),n)
    {
        int st,en;
        scanf("%d%d",&st,&en);
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        printf("%d\n",bfs(st,en));
    }
    return 0;
}