HDU 1548

A strange lift

[align=center]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15642    Accepted Submission(s): 5853
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[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"?

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[align=left]
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[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.
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[align=left]Output[/align]

[align=left]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]
 

[align=left]Sample Input[/align]

5 1 5
3 3 1 2 5
0

 

[align=left]Sample Output[/align]

3

题意是通过既定的规则从某一层到达另一层，这道题的特点在于每一层都有自己的步数，所以在广度搜索时要记得对应层数，并且限定范围防止超出输入的层数界限，我预先将所有的位置步数初始化成-1所以不注意输入的层数限制会出问题（没注意这点当成越界问题结果WA了），如果使用其他方式来记录就可以无视这点，一旦目标层数有了步数记录，就可以结束搜索了。

#include<iostream>
#include<cstdio>
#include<queue>
#include<memory.h>
using namespace std;

int N,K,L,D;
int d[500];
int a[410];
queue<int> M;
void bfs(int n){
while(!M.empty()) M.pop();

M.push(n);
while(!M.empty()&&a[K]==-1){
int nx1,nx2,cx;
cx=M.front();
M.pop();
D=a[cx]+1;
nx1=cx+d[cx];
nx2=cx-d[cx];
if(nx1>0&&nx1<=K&&a[nx1]==-1){
M.push(nx1);
a[nx1]=D;
}
if(nx2>0&&nx2<=K&&a[nx2]==-1){
M.push(nx2);
a[nx2]=D;
}
}
}
int main()
{
while(scanf("%d",&L)!=EOF){
if(L==0) break;

scanf("%d%d",&N,&K);
memset(a,-1,sizeof(a));
a
=0;
for(int i=1;i<=L;i++) scanf("%d",&d[i]);

if(N==K){
printf("0\n");
}else{
bfs(N);
printf("%d\n",a[K]);
}
}
return 0;
}