题目

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

题解

BFS题目
直接套用模板即可

其中有一点是如果"DOWN"到负数楼层，则不下降层数，而非降至1楼；"UP"同理（我觉得没人会读错吧……）

代码

/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o<n;o++)

const int maxn = 205;
int k[maxn];
int N;

int BFS(int s,int v) {
    if(s == v)
        return 0;

    queue<int> Q;
    bool visited[maxn];
    memset(visited,false,sizeof(visited));
    int dis[maxn];
    memset(dis,0,sizeof(dis));

    Q.push(s);
    visited[s] = true;
    while(!Q.empty()) {
        int th = Q.front();
        Q.pop();

        //达到终点
        if(th == v)
            break;

        //拓展节点
        int next;
        for(int i = -1;i == -1 || i == 1;i += 2) {
            next = th + i * k[th];
            if(next > N || next <= 0)
                continue;
            if(!visited[next]) {
                Q.push(next);
                visited[next] = true;
                dis[next] = dis[th] + 1;
            }
        }

    }

    if(dis[v])
        return dis[v];
    else
        return -1;
}

bool Do() {
    int s,v;
    if(scanf("%d%d%d",&N,&s,&v),N == 0)
        return false;
    REP(N)
        scanf("%d",&k[o + 1]);
    printf("%d\n",BFS(s,v));
    return true;
}

int main() {
    while(Do());
    return 0;
}