Subway POJ - 2502 最短路

Posted 2021-01-10 ttttttttrx

题意：给出地铁线  起点和 终点  坐地铁速度为v2  走路为v1 求起点到终点的最短距离  （答案需要四舍五入这里坑了好久） 

 拿给出的地铁站点 和起点终点建边即可  然后跑个迪杰斯特拉

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 using namespace std;
 6 const double v1=10000.0/60;
 7 const double v2=40000.0/60;
 8 int n;
 9 const int maxn=300+5;
10 double dist[maxn];
11 double cost[maxn][maxn];
12 int vis[maxn];
13 const double INF=1e30;
14 
15 struct Node{
16     double x,y;
17 }node[maxn];
18 
19 double dis(const Node&a,const Node&b){
20     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
21 }
22 
23 void Dijkstra(){
24     for(int i=1;i<=n;i++){
25         dist[i]=INF;
26     }
27     memset(vis,0,sizeof(vis));
28     dist[1]=0;
29     for(int j=0;j<n;j++){
30         int k=-1;
31         double minnum=INF;
32         for(int i=1;i<=n;i++){
33             if(!vis[i]&&dist[i]<minnum){
34                 minnum=dist[i];
35                 k=i;
36             }
37         }
38         if(k==-1)break;
39     vis[k]=1;
40     for(int i=1;i<=n;i++){
41         if(!vis[i]&&dist[k]+cost[k][i]<dist[i]){
42             dist[i]=dist[k]+cost[k][i];
43     }
44     }
45 }
46 }
47 int main(){
48     while(scanf("%lf%lf%lf%lf",&node[1].x,&node[1].y,&node[2].x,&node[2].y)==4){
49         n=2;
50     int cnt1=3;
51         for(int i=1;i<maxn;i++)
52             for(int j=1;j<maxn;j++)
53                 if(i!=j)cost[i][j]=INF;
54                 else cost [i][j]=0;
55         int x,y;
56         bool ok=0;
57         
58         while(scanf("%d%d",&x,&y)==2){
59             if(x==-1&&y==-1){
60             ok=0;
61                 continue;
62             }
63             n++;
64             node[n].x=x;
65             node[n].y=y;
66             if(ok){
67                 cost[n][n-1]=cost[n-1][n]=min(cost[n][n-1],dis(node[n],node[n-1])/v2);
68             }
69             ok=1;
70         }
71         for(int i=1;i<=n;i++)
72             for(int j=1;j<=n;j++){
73                cost[i][j]=cost[j][i]=min(cost[i][j],dis(node[i],node[j])/v1);
74             }
75         Dijkstra();
76     cout << int( dist[2] + 0.5 );
77     }
78     return 0;
79 }