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题目链接:

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

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

Problem solving report:

Description: 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？

Problem solving: 广搜，标准的BFS。。。

#include 
   
    #include 
    
     #include 
     
      using namespace std;int dir[2] = {1, -1};int s, e, n, st[205], vis[205];struct edge {    int lift, step;}t, p[201];int BFS(){    queue 
      
        Q;    memset(vis, 0, sizeof(vis));    vis[s] = 1;    Q.push((edge){s, 0});    while (!Q.empty())    {        t = Q.front();        Q.pop();        if (!(t.lift - e))            return t.step;        for (int i = 0; i < 2; i++)        {            int tx = t.lift + dir[i] * st[t.lift];            if (tx > 0 && tx <= n && !vis[tx])            {                vis[t.lift] = 1;                Q.push((edge){tx, t.step + 1});            }        }    }    return -1;}int main(){    while (scanf("%d", &n), n)    {        scanf("%d%d", &s, &e);        for (int i = 1; i <= n; i++)            scanf("%d", &st[i]);        printf("%d\n", BFS());    }    return 0;}
      
     
    
   

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