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

 

Sample Output

 

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最短路的模板题，代码还是很好理解的。

 

 #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 ;

 }

 #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();

     }

 }