hdu 1548 A strange lift（迪杰斯特拉，邻接表）

A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18723    Accepted Submission(s):
6926

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

 

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最短路的模板题，代码还是很好理解的。

 

题意：一个特别的电梯，按up可升上k[i]层，到大i+k[i]层，down则到达i-k[i]层，最高不能超过n，最低不能小于1，给你一个起点和终点，问最少可以按几次到达目的地。在一个N层高的楼有一个奇怪的电梯，在每一层只能上升或下降一个特定的层数，中间不会停止，在给定的条件下，问能不能到达指定楼层，可以到达的话返回转操作次数，不可以的话返回-1.

附上代码：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #define M 205
 #define MAX 0x3f3f3f3f
 using namespace std;
 int map[M][M],vis[M],dis[M];
 int main()
 {
     int n,a,b,i,j,s;
     while(~scanf("%d",&n)&&n)
     {
         scanf("%d%d",&a,&b);
         memset(vis,,sizeof(vis));
         memset(dis,,sizeof(dis));
         for(i=; i<=n; i++)
             for(j=; j<=n; j++)
             {
                 if(i==j)
                     map[i][j]=;  //同一个地方距离为0
                 else
                     map[i][j]=MAX;
             }
         for(i=; i<=n; i++)
         {
             scanf("%d",&s);
             if(i+s<=n)       //标记这一层电梯可以去的楼层
                 map[i][s+i]=;
             if(i-s>=)
                 map[i][i-s]=;
         }
         vis[a]=;    //起点已走过
         for(i=; i<=n; i++)
             dis[i]=map[a][i];   //初始化距离为每个点到起点的距离
         int min,k,t;
         for(i=; i<=n; i++)
         {
             min=MAX;
             for(j=; j<=n; j++)
                 if(!vis[j]&&dis[j]<min)  //每次都找离终点最近的点
                 {
                     min=dis[j];
                     t=j;
                 }
             vis[t]=;               //标记为已经找过此点
             for(j=; j<=n; j++)
                 if(!vis[j]&&map[t][j]<MAX)   //从最近的点到下一个点的距离与初始距离进行比较
                     if(dis[j]>dis[t]+map[t][j])
                         dis[j]=dis[t]+map[t][j];
         }
         if(dis[b]<MAX)
         printf("%d\n",dis[b]);
         else           //不能到 则输出-1
         printf("-1\n");
     }
     return ;
 }

邻接表代码：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <queue>
 #define inf 0x3f3f3f3f
 using namespace std;
 struct Edge
 {
     int from,to,val,next;
 }edge[];
 int tol,s,t,n;
 int dis[];
 bool vis[];
 int head[];

 void init()
 {
     tol=;
     memset(head,-,sizeof(head));
 }

 void addEdge(int u,int v)
 {
     edge[tol].from=u;
     edge[tol].to=v;
     edge[tol].val=;
     edge[tol].next=head[u];
     head[u]=tol++;
 }

 void getmap()
 {
     int x;
     for(int i=;i<=n;i++)
     {
         scanf("%d",&x);
         if(i-x>=) addEdge(i,i-x);
         if(i+x<=n) addEdge(i,i+x);
     }
     memset(vis,false,sizeof(vis));
     memset(dis,inf,sizeof(dis));
 }

 void spfa()
 {
     queue<int>q;
     q.push(s);
     vis[s]=true;
     dis[s]=;
     while(!q.empty())
     {
         int u=q.front();
         q.pop();
         vis[u]=false;
         for(int i=head[u];i!=-;i=edge[i].next)
         {
             int v=edge[i].to;
             if(dis[v]>dis[u]+edge[i].val)
             {
                 dis[v]=dis[u]+edge[i].val;
                 if(!vis[v])
                 {
                     vis[v]=true;
                     q.push(v);
                 }
             }
         }
     }
     if(dis[t]<inf)
     printf("%d\n",dis[t]);
     else
     printf("-1\n");
     return;
 }

 int main()
 {
     int i,j;
     while(~scanf("%d",&n)&&n)
     {
         scanf("%d%d",&s,&t);
         init();
         getmap();
         spfa();
     }
 }