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））
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <algorithm> 5 # include <cmath> 6 # define LL long long 7 using namespace std ; 8 9 const int MAXN=300; 10 const int INF=0x3f3f3f3f; 11 int n ; 12 bool vis[MAXN]; 13 int cost[MAXN][MAXN] ; 14 int lowcost[MAXN] ; 15 int pre[MAXN]; 16 void Dijkstra(int beg) 17 { 18 for(int i=0;i<n;i++) 19 { 20 lowcost[i]=INF;vis[i]=false;pre[i]=-1; 21 } 22 lowcost[beg]=0; 23 for(int j=0;j<n;j++) 24 { 25 int k=-1; 26 int Min=INF; 27 for(int i=0;i<n;i++) 28 if(!vis[i]&&lowcost[i]<Min) 29 { 30 Min=lowcost[i]; 31 k=i; 32 } 33 if(k==-1) 34 break ; 35 vis[k]=true; 36 for(int i=0;i<n;i++) 37 if(!vis[i]&&lowcost[k]+cost[k][i]<lowcost[i]) 38 { 39 lowcost[i]=lowcost[k]+cost[k][i]; 40 pre[i]=k; 41 } 42 } 43 44 } 45 46 47 48 int main () 49 { 50 // freopen("in.txt","r",stdin) ; 51 52 while (scanf("%d" , &n ) !=EOF) 53 { 54 if (n==0) 55 break ; 56 int i , j ; 57 for (i = 0 ; i < n ; i++) 58 for (j = 0 ; j < n ; j++) 59 cost[i][j] = INF ; 60 int s , e , t; 61 scanf("%d %d" , &s , &e) ; 62 for (i = 0 ; i < n ; i++) 63 { 64 scanf("%d" , &t) ; 65 if (i + t <= n-1) 66 cost[i][i+t] = 1 ; 67 if (i - t >= 0) 68 cost[i][i-t] = 1 ; 69 } 70 Dijkstra(s-1) ; 71 if (lowcost[e-1] != INF) 72 printf("%d " , lowcost[e-1]) ; 73 else 74 printf("-1 ") ; 75 } 76 77 return 0 ; 78 }
堆优化：
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <algorithm> 5 # include <cmath> 6 # include <queue> 7 # define LL long long 8 using namespace std ; 9 10 const int INF=0x3f3f3f3f; 11 const int MAXN=300; 12 struct qnode 13 { 14 int v; 15 int c; 16 qnode(int _v=0,int _c=0):v(_v),c(_c){} 17 bool operator <(const qnode &r)const 18 { 19 return c>r.c; 20 } 21 }; 22 struct Edge 23 { 24 int v,cost; 25 Edge(int _v=0,int _cost=0):v(_v),cost(_cost){} 26 }; 27 vector<Edge>E[MAXN]; 28 bool vis[MAXN]; 29 int dist[MAXN]; 30 int n ; 31 void Dijkstra(int start)//点的编号从1开始 32 { 33 memset(vis,false,sizeof(vis)); 34 for(int i=1;i<=n;i++)dist[i]=INF; 35 priority_queue<qnode>que; 36 while(!que.empty())que.pop(); 37 dist[start]=0; 38 que.push(qnode(start,0)); 39 qnode tmp; 40 while(!que.empty()) 41 { 42 tmp=que.top(); 43 que.pop(); 44 int u=tmp.v; 45 if(vis[u])continue; 46 vis[u]=true; 47 for(int i=0;i<E[u].size();i++) 48 { 49 int v=E[tmp.v][i].v; 50 int cost=E[u][i].cost; 51 if(!vis[v]&&dist[v]>dist[u]+cost) 52 { 53 dist[v]=dist[u]+cost; 54 que.push(qnode(v,dist[v])); 55 } 56 } 57 } 58 } 59 void addedge(int u,int v,int w) 60 { 61 E[u].push_back(Edge(v,w)); 62 } 63 64 int main () 65 { 66 // freopen("in.txt","r",stdin) ; 67 int m ; 68 while (scanf("%d" , &n) !=EOF) 69 { 70 if (n==0 ) 71 break ; 72 int u , v , w ; 73 int i , j ; 74 for(i=1;i<=n;i++) 75 E[i].clear(); 76 77 int s , e , t; 78 scanf("%d %d" , &s , &e) ; 79 for (i = 1 ; i <= n ; i++) 80 { 81 scanf("%d" , &t) ; 82 if (i + t <= n) 83 addedge(i,i+t,1) ; 84 if (i - t >= 1) 85 addedge(i,i-t,1) ; 86 } 87 Dijkstra(s) ; 88 if (dist[e] != INF) 89 printf("%d " , dist[e]) ; 90 else 91 printf("-1 ") ; 92 } 93 94 return 0 ; 95 }