A strange lift

D - A strange lift

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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 

 

PS：题目意思是给出n,a,b分别代表电梯有n层，你现在在第a层，想到第b层

接下来给出n个数，代表第i层能向上或向下走j层

比如题目给的例子

电梯有5层，你在第一层，想到第五层

第一层为3，即可向上或向下走3层，中间不能停，而且层数不能低于1或高于5，

要求出从1到5最小次数

题目为1向上走3，到了第4层，向下走2，到了第2层，再向上走3，到终点，共3步

注意a==b，结果为0

代码：

#include<iostream>#include<stdio.h>#include<string.h>#include<string>#include<algorithm>#include<queue>#include<math.h>using namespace std;int flo[250];bool flag[250];int n, a, b;struct F{int i;int c;}f;queue<F> fs;int dis[2] = { -1, 1 };void bfs(){while (!fs.empty()){F f1 = fs.front();fs.pop();for (int i = 0; i < 2; ++i){int x = f1.i + flo[f1.i] * dis[i];if (x == b){cout << f1.c + 1 << endl;while (!fs.empty()){fs.pop();}return;}if (x <= n && x >= 1 && !flag[x]){f.i = x;flag[x] = true;f.c = f1.c + 1;fs.push(f);}}}cout << -1 << endl;}int main(){while (cin >> n && n){memset(flo, 0, sizeof(flo));memset(flag, 0, sizeof(flag));cin >> a >> b;for (int i = 1; i <= n; ++i){cin >> flo[i];}if (a == b){cout << 0 << endl;}else{f.i = a;f.c = 0;flag[a] = true;fs.push(f);bfs();}}return 0;}