HDU--杭电--1548--A strange lift--最短路

A strange lift

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

Total Submission(s): 9377    Accepted Submission(s): 3555

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：输入t,n,m分别代表楼梯层数、起点和终点，接下来是t层楼的数字，数字的意义是电梯从这层楼只能向上或者向下走的固定层数，但是不会低于1楼高于t楼，如：1楼数字是2，向上只能到3楼，向下不能走。

#include 
#include 
#include 
#include 
#define Max 0xfffffff
using namespace std;
queue q,qq;
int t,n,m,s[222],d[222];    //s是伪地图，d是电梯运行次数
bool v[222];    //记录是否已入队
void add(int a,int b,int x) //松弛操作
{
    if(d[a]>b)
    {
        d[a]=b;
        if(v[x])    //如果在队外就入队
        {
            q.push(x);
            v[x]=false;
        }
    }
}
void SPFA()
{
    int k;
    while(!q.empty())
    {
        k=q.front();q.pop();
        v[k]=true;
        if(k+s[k]<=t)   //只有上楼梯和下楼梯两种，只要判断没出界就好
        add(k+s[k],d[k]+1,k+s[k]);  //k+s[k]是向上走到达的楼层，跟电梯运行次数比
        if(k-s[k]>0)
        add(k-s[k],d[k]+1,k-s[k]);
    }
}
int main (void)
{
    int i,j,k,l;
    while(~scanf("%d",&t),t)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=t;i++)
        {
            scanf("%d",&s[i]);
            v[i]=true;
            d[i]=Max;
        }
        q=qq;
        q.push(n);
        d[n]=0;
        v[n]=false;
        SPFA();
        printf("%d\n",d[m]