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=300;
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=0;i<n;i++)
    {
        lowcost[i]=INF;vis[i]=false;pre[i]=-1;
    }
    lowcost[beg]=0;
    for(int j=0;j<n;j++)
    {
        int k=-1;
        int Min=INF;
        for(int i=0;i<n;i++)
            if(!vis[i]&&lowcost[i]<Min)
            {
                Min=lowcost[i];
                k=i;
            }
            if(k==-1)
                break ;
            vis[k]=true;
            for(int i=0;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==0)
            break ;
        int i , j ;
        for (i = 0 ; i < n ; i++)
            for (j = 0 ; j < n ; j++)
               cost[i][j] = INF ;
        int s , e , t;
        scanf("%d %d" , &s , &e) ;
        for (i = 0 ; i < n ; i++)
        {
            scanf("%d" , &t) ;
            if (i + t <= n-1)
                cost[i][i+t] = 1 ;
            if (i - t >= 0)
                cost[i][i-t] = 1 ;
        }
        Dijkstra(s-1) ;
        if (lowcost[e-1] != INF)
            printf("%d\n" , lowcost[e-1]) ;
        else
            printf("-1\n") ;
    }

    return 0 ;
}

 

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=300;
struct qnode
{
    int v;
    int c;
    qnode(int _v=0,int _c=0):v(_v),c(_c){}
    bool operator <(const qnode &r)const
    {
        return c>r.c;
    }
};
struct Edge
{
    int v,cost;
    Edge(int _v=0,int _cost=0):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=1;i<=n;i++)dist[i]=INF;
    priority_queue<qnode>que;
    while(!que.empty())que.pop();
    dist[start]=0;
    que.push(qnode(start,0));
    qnode tmp;
    while(!que.empty())
    {
        tmp=que.top();
        que.pop();
        int u=tmp.v;
        if(vis[u])continue;
        vis[u]=true;
        for(int i=0;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==0 )
            break ;
        int u , v , w ;
        int i , j ;
        for(i=1;i<=n;i++)
            E[i].clear();

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

    return 0 ;
}