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 ;
}