最短路径__Subway( Poj 2502 )

Description

You have just moved from a quiet Waterloo neighbourhood to a big, noisy city. Instead of getting to ride your bike to school every day, you now get to walk and take the subway. Because you don't want to be late for class, you want to know how long it will take you to get to school. 
You walk at a speed of 10 km/h. The subway travels at 40 km/h. Assume that you are lucky, and whenever you arrive at a subway station, a train is there that you can board immediately. You may get on and off the subway any number of times, and you may switch between different subway lines if you wish. All subway lines go in both directions.

Input

Input consists of the x,y coordinates of your home and your school, followed by specifications of several subway lines. Each subway line consists of the non-negative integer x,y coordinates of each stop on the line, in order. You may assume the subway runs in a straight line between adjacent stops, and the coordinates represent an integral number of metres. Each line has at least two stops. The end of each subway line is followed by the dummy coordinate pair -1,-1. In total there are at most 200 subway stops in the city.

Output

Output is the number of minutes it will take you to get to school, rounded to the nearest minute, taking the fastest route.

Sample Input

0 0 10000 1000
0 200 5000 200 7000 200 -1 -1 
2000 600 5000 600 10000 600 -1 -1

Sample Output

21

题目大意：首先给你两个坐标，一个是你家里的坐标，一个是你学校的坐标，然后接下来有若干条地铁线，每条地铁线上有若干个站点，给出每个站点的坐标，有这些点，这些点当中有距离，这个距离的单位是米，现在告诉你走路是10km/h,做地铁的话是40km/h，问你从家里到学校所花费的最短时间（分钟）

思路：这道题主要是考建边,对于地铁方式每相邻两个站建一条边,对于步行方式每两个站都可以建一条边,建边用时间作为权值

#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<string.h>
#include<iostream>
#include<cmath>
#include<stdlib.h>
#define N 310
#define INF 0x7fffffff
using namespace std;
struct edge
{
    double x,y;
}team[N];
int visit[N],n;
double mpt[N][N],dis[N];
double Dis(int a,int b)
{
    double x = team[a].x - team[b].x;
    double y = team[a].y - team[b].y;
    return sqrt(x*x+y*y);
}
void Add_edge(int a,int b,double t)
{
    if(mpt[a][b] > t) mpt[a][b] = t;
    if(mpt[b][a] > t) mpt[b][a] = t;
}

void Dijkstra()
{
    for(int i = 1 ; i <= n ; i ++) dis[i] = mpt[1][i];
    memset(visit,0,sizeof(visit));
    visit[1] = 1;
    int i,j,Min,Minj;
    for(i = 0 ; i < n ; i ++)
    {

        Min = INF;Minj = -1;
        for(j = 1 ; j <= n ; j ++)
        {
            if(visit[j])continue;
            if(dis[j] < Min)
            {
                Min = dis[j];
                Minj = j;
            }
        }
        if(Minj == -1) break;
        visit[Minj] = 1;
        for(j = 1; j <= n ; j ++)
        {
            if(dis[j] > dis[Minj] + mpt[Minj][j]) dis[j] = dis[Minj] + mpt[Minj][j];
        }
    }
}


int main()
{
    int i,j,k,m;
    double x,y;
    double v1=10000.0/60;  //步行
    double v2=40000.0/60;  //地铁
    n =  2;m = 3;

    for(i = 0 ; i < N ; i ++)
    {
        for( j = 0 ; j < N ; j ++)
        {
            if(i == j) mpt[i][j] = 0;
            else mpt[i][j] = INF;
        }
    }
    scanf("%lf %lf %lf %lf",&team[1].x,&team[1].y,&team[2].x,&team[2].y);
    while(scanf("%lf %lf",&x,&y)!=EOF)
    {
        if( x == -1 && y == -1 )
        {
            m = n + 1;
            continue;
        }
        n++;
        team[n].x = x;team[n].y = y;
        if( m != n)
        {
            Add_edge(n-1,n,Dis(n-1,n)/v2);
        }
    }

    for(i = 1; i <= n; i ++)
    {
        for(j = i + 1; j <= n ; j ++)
        {
            Add_edge(i,j,Dis(i,j)/v1);
        }
    }
    Dijkstra();
    printf("%d\n",(int)(dis[2]+0.5));

}