A strange lift
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15570 Accepted Submission(s): 5832
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"?
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.
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
Recommend
电梯仅仅有两个方向，向上或者向下。既然是求最短路径。也就用到dijkstra算法（无负权值）。
仅仅要能想到怎样构造算法即可。假设自己的算法，却不知道怎样用来解题，也都是没用的。
在这里我想给大家说一下。
在以后做题的过程中，不要仅仅看别人的代码，要看思想，别人为什么这样写。然后依据自己想象的思想写一遍代码。写的过程中不要
看别人的代码。即使不正确也无所谓，这样印象最深，以后也就随手敲来、
详细还是代码里面见：
#include <stdio.h>
#include<string.h>
#include <queue>
using namespace std;
struct node
{
int pos,t;
friend bool operator<(node a,node b)
{
return a.t>b.t;
}
};
priority_queue<node>s;
int lift[205],vis[205],n;
int dijkstra(int st,int ed)
{
node temp,temp1;
int flag=0;
temp.pos=st,temp.t=0;
s.push(temp);
while(!s.empty())
{
temp1=temp=s.top(),s.pop();
vis[temp.pos]=1;
if(temp.pos==ed)
{
flag=1;
break;
}
temp.pos=temp1.pos-lift[temp1.pos];//
temp.t=temp1.t+1;
if(temp.pos>=1&&temp.pos<=n&&!vis[temp.pos])
s.push(temp);
temp.pos=temp1.pos+lift[temp1.pos];
temp.t=temp1.t+1;
if(temp.pos>=1&&temp.pos<=n&&!vis[temp.pos])
s.push(temp);//和以往的代码不同的也就这个地方。曾经做的要么是个矩阵，要么是个树，如今就两个方向了。。推断电梯的这两个方向，进队列即可了
}
if(flag)
return temp.t;
else
return -1;
}
int main()
{
int st,ed;
while(scanf("%d",&n)!=EOF)
{
if(n==0)
break;
scanf("%d %d",&st,&ed);
for(int i=1;i<=n;i++)
scanf("%d",&lift[i]);
memset(vis,0,sizeof(vis));
while(!s.empty())
s.pop();
printf("%d
",dijkstra(st,ed));
}
return 0;
}