HDU - 1548

A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31944    Accepted Submission(s): 11441

 

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】

     给你一个数组，ki代表在第i层，电梯只能上或者下k楼，楼高为n,电梯只能到达1-n楼，电梯初始在A楼，求到达b楼的最小步数

 

【思路】

    明显是一道简单的bfs，一开始写的时候认为不需要vis数组标记，结果就错了！ 错了！！！！以后要记着Bfs都需要标记数组来进行标记，对于这道题，你走到i楼，就一定会有上ki或者下ki楼是它的最佳方案，不存在走过i楼后，需要再次走到i楼的情况，所以需要用VIS数组去标记是否到达过i楼

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
int k[20009];
int pos[200009],sum[2000009];
int vis[310000],n;
void bfs(int a,int b){
	int head = 0,tail = 1;
	pos[1] = a;
	vis[a] = 1;
	while (head < tail){
		head++;
		if (pos[head] == b){
			printf("%d\n",sum[head]);
			return;
		}
		if (pos[head]+k[pos[head]]<=n && vis[pos[head]+k[pos[head]]] == 0){
			tail++;
			pos[tail] = pos[head]+k[pos[head]];
			sum[tail] = sum[head]+1;
			vis[pos[head]+k[pos[head]]] = 1;
		}
		if (pos[head]-k[pos[head]]>0 && vis[pos[head]-k[pos[head]]] == 0){
			tail++;
			pos[tail] = pos[head]-k[pos[head]];
			sum[tail] = sum[head]+1;
			vis[pos[head]-k[pos[head]]] = 1;
		}
	}
	printf("-1\n");
	return ;
}
int main ()
{
	int a,b;
	while (~scanf("%d",&n)){
		if (n == 0) break;
		scanf("%d%d",&a,&b);
		memset(pos,0,sizeof(pos));
		memset(k,0,sizeof(k));
		memset(sum,0,sizeof(sum));
		memset(vis,0,sizeof(vis));
		for (int i = 1; i <= n; i++)
			scanf("%d",&k[i]);
		bfs(a,b);
	}
}