hdu 1548 升降梯

题目大意：有一个升降机，它有两个按钮UP和DOWN，给你一些数i表示层数，并且每层对应的Ki，如果按UP按钮，会从第i层升到第i+Ki层；如果按了DOWN则会从第i层降到第i-Ki层；并规定能到的层数为1到N，现在的要求就是给你N,A,B和一串数K1到Kn，问你从A到B，至少按几下按钮。

构造一个图，边的权值为1

Sample Input
5 1 5 //层数 起点 终点
3 3 1 2 5
0

Sample Output
3

Dijkstra：（O（n^2））

 # include <iostream>
 # include <cstdio>
 # include <cstring>
 # include <algorithm>
 # include <cmath>
 # define LL long long
 using namespace std ;

 const int MAXN=;
 const int INF=0x3f3f3f3f;
 int n ;
 bool vis[MAXN];
 int cost[MAXN][MAXN] ;
 int lowcost[MAXN] ;
 int pre[MAXN];
 void Dijkstra(int beg)
 {
     for(int i=;i<n;i++)
     {
         lowcost[i]=INF;vis[i]=false;pre[i]=-;
     }
     lowcost[beg]=;
     for(int j=;j<n;j++)
     {
         int k=-;
         int Min=INF;
         for(int i=;i<n;i++)
             if(!vis[i]&&lowcost[i]<Min)
             {
                 Min=lowcost[i];
                 k=i;
             }
             if(k==-)
                 break ;
             vis[k]=true;
             for(int i=;i<n;i++)
                 if(!vis[i]&&lowcost[k]+cost[k][i]<lowcost[i])
                 {
                     lowcost[i]=lowcost[k]+cost[k][i];
                         pre[i]=k;
                 }
     }

 }

 int main ()
 {
    // freopen("in.txt","r",stdin) ;

     while (scanf("%d" , &n ) !=EOF)
     {
         if (n==)
             break ;
         int i , j ;
         for (i =  ; i < n ; i++)
             for (j =  ; j < n ; j++)
                cost[i][j] = INF ;
         int s , e , t;
         scanf("%d %d" , &s , &e) ;
         for (i =  ; i < n ; i++)
         {
             scanf("%d" , &t) ;
             if (i + t <= n-)
                 cost[i][i+t] =  ;
             if (i - t >= )
                 cost[i][i-t] =  ;
         }
         Dijkstra(s-) ;
         if (lowcost[e-] != INF)
             printf("%d\n" , lowcost[e-]) ;
         else
             printf("-1\n") ;
     }

     return  ;
 }

堆优化：

 # include <iostream>
 # include <cstdio>
 # include <cstring>
 # include <algorithm>
 # include <cmath>
 # include <queue>
 # define LL long long
 using namespace std ;

 const int INF=0x3f3f3f3f;
 const int MAXN=;
 struct qnode
 {
     int v;
     int c;
     qnode(int _v=,int _c=):v(_v),c(_c){}
     bool operator <(const qnode &r)const
     {
         return c>r.c;
     }
 };
 struct Edge
 {
     int v,cost;
     Edge(int _v=,int _cost=):v(_v),cost(_cost){}
 };
 vector<Edge>E[MAXN];
 bool vis[MAXN];
 int dist[MAXN];
 int n ;
 void Dijkstra(int start)//点的编号从1开始
 {
     memset(vis,false,sizeof(vis));
     for(int i=;i<=n;i++)dist[i]=INF;
     priority_queue<qnode>que;
     while(!que.empty())que.pop();
     dist[start]=;
     que.push(qnode(start,));
     qnode tmp;
     while(!que.empty())
     {
         tmp=que.top();
         que.pop();
         int u=tmp.v;
         if(vis[u])continue;
         vis[u]=true;
         for(int i=;i<E[u].size();i++)
         {
             int v=E[tmp.v][i].v;
             int cost=E[u][i].cost;
             if(!vis[v]&&dist[v]>dist[u]+cost)
             {
                 dist[v]=dist[u]+cost;
                 que.push(qnode(v,dist[v]));
             }
         }
     }
 }
 void addedge(int u,int v,int w)
 {
     E[u].push_back(Edge(v,w));
 }

 int main ()
 {
   //  freopen("in.txt","r",stdin) ;
     int m ;
     while (scanf("%d" , &n) !=EOF)
     {
         if (n== )
             break ;
         int u , v , w ;
         int i , j ;
         for(i=;i<=n;i++)
             E[i].clear();

         int s , e , t;
         scanf("%d %d" , &s , &e) ;
         for (i =  ; i <= n ; i++)
         {
             scanf("%d" , &t) ;
             if (i + t <= n)
                 addedge(i,i+t,) ;
             if (i - t >= )
                 addedge(i,i-t,) ;
         }
         Dijkstra(s) ;
         if (dist[e] != INF)
             printf("%d\n" , dist[e]) ;
         else
             printf("-1\n") ;
     }

     return  ;
 }