HDU 1548 A strange lift

                                                           A strange lift

[align=left]Problem Description[/align]
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"?

 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
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".
 

[align=left]Sample Input[/align]

5 1 5
3 3 1 2 5
0

[align=left]Sample Output[/align]

3

 
[align=left]题意：输入N，表示有N层楼，输入A，表示初始在A层楼，输入B，表示目的地在B层楼，然后输入N个数，表示在第i层楼可以上下a[i]层楼，求至少需要几步才能到达B层楼。[/align]
[align=left]
[/align]
[align=left]分析：感觉是一道挺简单的BFS的题目，然而TLE了好多发，后来发现一层楼只能经过一次，否则的话会出现重复的情况，因此需要标记每次到达了哪层楼。[/align]
[align=left]
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[align=left]代码：[/align]

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int n,s,e;
int a[250];
int id[250];
int ans;
struct node
{
int x;
int step;
};
void bfs()
{
node t;
t.x=s;
t.step=0;
id[s]=1;
queue<node> q;
q.push(t);
while(q.empty()==0)
{
node t=q.front();
q.pop();
if(t.x==e)
{
ans=t.step;
return ;
}
node t1,t2;
t1.x=t.x+a[t.x];
if(t1.x>=1&&t1.x<=n&&id[t1.x]==0)
{
id[t1.x]=1;
t1.step=t.step+1;
q.push(t1);
}
t2.x=t.x-a[t.x];
if(t2.x>=1&&t2.x<=n&&id[t2.x]==0)
{
t2.step=t.step+1;
id[t2.x]=1;
q.push(t2);
}
}
}
int main()
{
while(~scanf("%d",&n)&&n)
{
scanf("%d%d",&s,&e);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
ans=-1;
memset(id,0,sizeof(id));
bfs();
if(ans==-1)
printf("-1\n");
else
printf("%d\n",ans);
}
return 0;
}

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1548