hdu 1548 A strange lift

最新推荐文章于 2021-12-18 21:01:38 发布

原创最新推荐文章于 2021-12-18 21:01:38 发布 · 182 阅读

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本文介绍了一个有趣的算法问题：在一栋特殊的大楼中乘坐只能上行或下行特定楼层数的电梯，如何从任意一层到达目标层并计算最少的操作次数。文章通过实例展示了使用Dijkstra算法解决该问题的过程。

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

这道题初始想法是用BFS，后来发现原来dijkstra也可以做。

具体做法就是将每一层电梯看作一个结点，如果层x有通向层y的路，则将x，y边的权值记为1，无路则为INF（构边时为单向边）然后dijkstra求从结点A到结点B的最短路径就

ok了。

#include<iostream>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f
int n,a,b;
int p[201];
int map[201][201];
int d[201];
int vis[201];
void dij(){
		memset(map,0x3f,sizeof(map));
		int i,j,v,min;
		for(i=1;i<=n;i++){
			map[i][i]=0;
			vis[i]=0;
			if(i+p[i]<=n)
				map[i][i+p[i]]=1;
			if(i-p[i]>=1)
				map[i][i-p[i]]=1;

		}
		for(i=1;i<=n;i++){
			d[i]=map[a][i];
		}
		for(i=1;i<=n;i++){
				min=INF;
				for(j=1;j<=n;j++){
					if(vis[j]==0&&d[j]<min){
						v=j;
						min=d[j];
					}

				}vis[v]=1;
				for(j=1;j<=n;j++){
					if(vis[j]==0&&d[j]>d[v]+map[v][j])
						d[j]=d[v]+map[v][j];
				}



		}if(d[b]<INF)
			cout<<d[b]<<endl;
		 else
			cout<<-1<<endl;





}
int main(){
	int i,j;
		while(cin>>n&&n!=0){
			cin>>a>>b;
			for(i=1;i<=n;i++){
				cin>>p[i];
			}
			dij();

		}



}