A strange lift

最新推荐文章于 2021-12-18 21:01:38 发布

DeDeWo

最新推荐文章于 2021-12-18 21:01:38 发布

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分类专栏： DFS and BFS 文章标签： lift button go input up output

本文链接：https://blog.youkuaiyun.com/aihahaheihei/article/details/7309662

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本文介绍了一个特殊的电梯系统，该系统允许乘客在指定楼层之间通过向上或向下按钮进行移动，每次移动距离固定。给定起始楼层、目标楼层及各楼层的特殊移动限制，本算法求解从起始楼层到达目标楼层所需的最少操作次数。通过广度优先搜索（BFS）算法，确保找到最短路径。

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B - A strange lift

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status

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 is on floor A,and you want to go to floor B,how many times at least he havt 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

Sample Output

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
struct node{
int x,num;
};
const int N=203;
bool visited[N+2];
int k[N+2];
int n;
int d;
int bfs(int v){
queue<node>q;
node ans,t;
int x;
ans.x=v;
ans.num=0;
q.push(ans);
visited[v]=true;
while(!q.empty()){
t=q.front();
q.pop();
x=t.x+k[t.x];
if(x<1 || x>n);
else if(!visited[x]){
visited[x]=true;
ans.x=x;
ans.num=t.num+1;
q.push(ans);
if(ans.x==d)
return ans.num;
}
x=t.x-k[t.x];
if(x<1 || x>n);
else if(!visited[x]){
visited[x]=true;
ans.x=x;
ans.num=t.num+1;
q.push(ans);
if(ans.x==d)
return ans.num;
}

}
return -1;
}
int main(){
int s;
while(cin>>n){
if(n==0) break;
cin>>s>>d;
for(int i=1;i<=n;i++)
cin>>k[i];
if(s==d){
cout<<0<<endl;
continue;
}
memset(visited,false,sizeof(visited));
cout<<bfs(s)<<endl;
}

//system("pause");
return 0;
}