ACM-BFS之A strange lift——hdu1548

A strange lift
题目： http://acm.hdu.edu.cn/showproblem.php?pid=1548
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

开始一些BFS练习了，先来一道简单的，这是一栋大楼，有一部很奇怪的电梯，电梯只有两个按钮，

UP和DOWN，每个楼层有一个数字num，如果你在第2层，你按UP，电梯会到达 num+2层，按DOWN亦然，

到达2-num层，求一个人从一个楼层要去另一个楼层，需要按几步？

（这个电梯够让人捉急的！）

当然如果碰到0层或者-1层都不能到达，碰到大于刚开始的n层的也不能到达，

如果永远到不了需要去的那一层，就输出-1了。

广搜，队列做，标准的BFS题目呀，很容易就AC了。。

#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
// st数组记录每个楼层能上行下行的数字，vis记录是否走过该楼层
int st[201],vis[201];
int start,finish,n;
// 该结构体记录到lift楼层用了几步
struct Elevator
{
    int lift,step;
}ele[201];

int bfs(void)
{
    Elevator t,k;
    queue <Elevator> q;
    memset(vis,0,sizeof(vis));

    // 第一个点入队列
    t.lift=start;
    t.step=0;
    vis[t.lift]=1;
    q.push(t);

    while(!q.empty())
    {
        t=q.front();
        q.pop();
        // 到达终点  直接返回步数
        if(t.lift==finish)  return t.step;

        // 从该楼层向下走
        k.lift=t.lift-st[t.lift];
        if(k.lift>0 && k.lift<=n && !vis[k.lift])
        {
            k.step=t.step+1;
            vis[t.lift]=1;
            q.push(k);
        }
        // 从该楼层向下走
        k.lift=t.lift+st[t.lift];
        if(k.lift>0 && k.lift<=n && !vis[k.lift])
        {
            vis[t.lift]=1;
            k.step=t.step+1;
            q.push(k);
        }
    }
    // 如果没有结果 -1
    return -1;
}

int main()
{
    int i;
    while(cin>>n)
    {
        if(!n)  break;

        cin>>start>>finish;
        for(i=1;i<=n;++i)   cin>>st[i];
        cout<<bfs()<<endl;
    }
    return 0;
}