Subway POJ

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const double v1=10000.0/60;
const double v2=40000.0/60;
int n;
const int maxn=300+5;
double dist[maxn];
double cost[maxn][maxn];
int vis[maxn];
const double INF=1e30;

struct Node{
    double x,y;
}node[maxn];

double dis(const Node&a,const Node&b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void Dijkstra(){
    for(int i=1;i<=n;i++){
        dist[i]=INF;
    }
    memset(vis,0,sizeof(vis));
    dist[1]=0;
    for(int j=0;j<n;j++){
        int k=-1;
        double minnum=INF;
        for(int i=1;i<=n;i++){
            if(!vis[i]&&dist[i]<minnum){
                minnum=dist[i];
                k=i;
            }
        }
        if(k==-1)break;
    vis[k]=1;
    for(int i=1;i<=n;i++){
        if(!vis[i]&&dist[k]+cost[k][i]<dist[i]){
            dist[i]=dist[k]+cost[k][i];
    }
    }
}
}
int main(){
    while(scanf("%lf%lf%lf%lf",&node[1].x,&node[1].y,&node[2].x,&node[2].y)==4){
        n=2;
    int cnt1=3;
        for(int i=1;i<maxn;i++)
            for(int j=1;j<maxn;j++)
                if(i!=j)cost[i][j]=INF;
                else cost [i][j]=0;
        int x,y;
        bool ok=0;
        
        while(scanf("%d%d",&x,&y)==2){
            if(x==-1&&y==-1){
            ok=0;
                continue;
            }
            n++;
            node[n].x=x;
            node[n].y=y;
            if(ok){
                cost[n][n-1]=cost[n-1][n]=min(cost[n][n-1],dis(node[n],node[n-1])/v2);
            }
            ok=1;
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++){
               cost[i][j]=cost[j][i]=min(cost[i][j],dis(node[i],node[j])/v1);
            }
        Dijkstra();
    cout << int( dist[2] + 0.5 );
    }
    return 0;
}