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

Dijkstra：

```#include<iostream>
using namespace std;
#define INF 999999
int 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：

```#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
#define maxn 250
bool visit[maxn];
int 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 n,start,end;
struct node
{
int pos,t;
}temp,p;
queue<node> q;
int BFS()
{
memset(visit,false,sizeof(visit));
while(!q.empty())
{
temp=q.front();
q.pop();
visit[temp.pos]=true;
if(temp.pos==end)
return temp.t;
int up=floor[temp.pos][0],down=floor[temp.pos][1];
if(up!=-1&&!visit[up])
{
p.pos=up;
p.t=temp.t+1;
q.push(p);
}
if(down!=-1&&!visit[down])
{
p.pos=down;
p.t=temp.t+1;
q.push(p);
}
}
return -1;
}
int main()
{
while(scanf("%d",&n),n)
{
scanf("%d %d",&start,&end);
memset(floor,-1,sizeof(floor));
while(!q.empty())
q.pop();
for(int i=1;i<=n;i++)
{
int t;
scanf("%d",&t);
temp.pos=i;
floor[i][0]=floor[i][1]=-1;
if(i+t<=n)
floor[i][0]=i+t;
if(i-t>=1)
floor[i][1]=i-t;
if(i==start)
{
temp.t=0;
q.push(temp);
}
}
temp=q.front();
printf("%d\n",BFS());
}
return 0;
}  ```

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 0x3f3f3f3f
int 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;
}```

原文作者：Suhx
原文地址: https://www.cnblogs.com/Su-Blog/archive/2012/08/27/2659214.html
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