poj 2502

Subway
                                        

> ***|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：
    给你俩个坐标第一个为家，第二个为学校。
    之后每一行给出一条地铁线路，给个线路相邻的俩个站点连通，每一条地铁线以-1 -1 结束。最多200个站点;
    步行速度10km/h 地铁40km/h;

难点是建图，用链接矩阵储存，如果是同一条线路且为上一个站点time=d /(40 * 1000 / 60);否则为d /(10 * 1000/60)，在nmap[i][j]比较取小

最后ans四舍五入;

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>

const int maxn=505;

using namespace std;

double nmap[maxn][maxn],d[maxn];

struct node{
    int x,y;
}e[maxn];

int vis[maxn],m,tot=0;

double len(node a,node b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void spfa(int st)
{
	memset(vis,0,sizeof(vis));
	memset(d,125,sizeof(d));
	queue<int>q;
	q.push(st);
	vis[st]=1;
	d[st]=0;
	while(q.size())
	{
		int now=q.front();
		q.pop();
		vis[now]=0;
		for(int i=1;i<=tot;i++)
		{
			double w=nmap[now][i];
			if(d[i]>d[now]+w)
			{
				d[i]=d[now]+w;
				if(vis[i]==0)
				{
					vis[i]=1;
					q.push(i);
				}
			}
		}
	}
}

int main(){
    int x,y,last=3;
    memset(nmap,125,sizeof(nmap));
    
    while(scanf("%d%d",&x,&y)!=EOF)
    {
        if(x==-1&&y==-1)
        {
            last=tot+1;
            continue;
        }
        else
        {
            e[++tot]=(node){x,y};
            for(int j=1;j<=tot;j++)
            {
                double tmp=len(e[tot],e[j]);
                if(j==tot-1&&j>=last) tmp/=(40.0*1000/60);
                else tmp/=(10.0*1000/60);
                nmap[j][tot]=nmap[tot][j]=min(nmap[tot][j],tmp);
            }
        }
    }

    spfa(1);
    printf("%d
",(int)floor(d[2]+0.5));
    return 0;
}