HDU 1548 A strange lift

A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18518    Accepted Submission(s): 6864

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 53 3 1 2 50

 

Sample Output

3

题意：你在电梯上，从a开始上升，终点到b。但是电梯每一层有个数字k，代表你只能上k或者下k层；但是不能低于1层，高于n层；

           问你需要调整按钮几次可以到终点；

PS：这里的技巧因为是次数，所以在满足情况的时候看，把可走的距离变为1就行了；

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define inf 9999999using namespace std ;int map[1111][1111],vis[1111],dis[1111],a, b, k[1111],n;void dijk(){for(int i = 1 ;i<=n;i++){dis[i]=map[a][i];}int mini , v ;for(int i = 1 ; i<=n;i++){mini= inf;for(int j=1 ;j <=n;j++){if(!vis[j]&&dis[j]<mini){mini=dis[j];v=j;}}vis[v]=1;for(int j= 1 ; j<=n;j++){if(!vis[j]&&dis[j]>dis[v]+map[v][j]){dis[j]=dis[v]+map[v][j];}}}}int main(){while(cin>>n&&n!=0){cin>>a>>b;for(int i =1 ;i<=n;i++){for(int j = 1 ;j<=n;j++){map[i][j]=inf;}map[i][i]=0;vis[i]=0;dis[i]=inf;}for(int i = 1 ; i<=n;i++){cin>>k[i];if(i+k[i]<=n){map[i][i+k[i]]=1;}if(i-k[i]>=1){map[i][i-k[i]]=1;}}dijk();if(dis[b]<inf){printf("%d\n",dis[b]);}else cout<<"-1"<<endl;}return 0 ;}