HDU 1458

 A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6706    Accepted Submission(s): 2499

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

 

 

Recommend

8600

 

 

 

可以用到的算法：Dijkstra、BFS、DP

题意:

有一个特别的电梯，第i层有一个对应的数字ki， 对于第i层按上升键up可升上到i+k[i]层，按下降键down到达i-k[i]层，到达的楼层最高不能超过n层，最低不能小于1层。给你一个起点A和终点B，问最少要按几次上升键或者下降键到达目的地。

 

思路一：
最短路：把每一层都看成一个节点，问题就可以变成求起点到终点的最短路径问题。
用Dijkstra算法和BFS算法都可以解。

 

Dijkstra：

 

#include<iostream>using namespace std;#define INF 999999int map[500][500];int dist[500],used[500];int n;void Dijkstra(int v){     int i,j,pos,min;    memset(used,0,sizeof(used));       for(i=1;i<=n;i++)        dist[i]=map[v][i];    used[v]=1;     dist[v]=0;//floor A==floor B(起点就是终点)的情况    for(j=1;j<n;j++)     {         pos=0;         min=100000000;         for(i=1;i<=n;i++)             if(dist[i]<min&&used[i]==0)            {                 pos=i;                min=dist[i];            }            used[pos]=1;             for(i=1;i<=n;i++)                if(dist[i]>dist[pos]+map[pos][i])                     dist[i]=dist[pos]+map[pos][i];     } } int main(){    int s,t,i,j;    int a[500];    while(cin>>n,n)    {        cin>>s>>t;        memset(map,INF,sizeof(map));        for(i=1;i<=n;i++)        {            cin>>a[i];            if(i+a[i]<=n)             map[i][i+a[i]]=1;//构图必须是单向边            if(i-a[i]>=1)             map[i][i-a[i]]=1;        }        Dijkstra(s);        if(dist[t]>INF)//If you can't reach floor B,printf "-1"            cout<<-1<<endl;        else        cout<<dist[t]<<endl;    }    return 0;}

 

 

BFS：

 

按 Ctrl+C 复制代码

按 Ctrl+C 复制代码

 

 

思路二：

DP: dp[i]表示从起点到第i层的按按钮的最少次数

 

 

DP：

 

#include<cstdio>#include<cstring>#define maxn 250#define INF 0x3f3f3f3fint dp[maxn],floor[maxn][2];//floor[i][0]表示第i层向上能到的楼层，floor[i][1]则表示向下能到的楼层int main(){    int n,a,b;    while(scanf("%d",&n),n)    {        scanf("%d %d",&a,&b);        memset(floor,-1,sizeof(floor));        for(int i=1;i<=n;i++)        {            int t;            scanf("%d",&t);            if(i+t<=n)                floor[i][0]=i+t;            if(i-t>=1)                floor[i][1]=i-t;        }        memset(dp,0x3f,sizeof(dp));        dp[a]=0;        while(true)        {            int num=0;//记录更新数据的次数            for(int i=1;i<=n;i++) if(dp[i]<INF)                for(int j=0;j<2;j++) if(floor[i][j]!=-1)                    if(dp[floor[i][j]]>dp[i]+1)                        dp[floor[i][j]]=dp[i]+1,num++;                    if(num==0)//无法继续更新                        break;        }        if(dp[b]==INF)            dp[b]=-1;        printf("%d\n",dp[b]);    }    return 0;}