POJ2502乘坐地铁上学

POJ2502

题目大意：给你一个二维地图，起点终点，以及多条地铁线路（只有相邻两站才可互通）中的站点坐标。步行以及坐地铁的速度；

思路：难点在于建图，一个多小时磨磨蹭蹭，都是因为思路没捋清楚~~，对于每一条地铁线，在输入站点坐标的过程中我们就可以建立图像了；输入完成后，遍历所有的点再建立，以步行为计量单位的图像。就ok啦，然后裸spfa就好了

#include <iostream>
#include <vector>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <queue>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 2e5 + 2e2;
struct Point
{
    double x,y;
    Point (double x = 0.0,double y = 0.0):x(x),y(y){}
}ps[maxn];
struct node
{
    int to,pre;
    double v;
}edge[maxn];
double d[maxn];
int vis[maxn];
int id[maxn];
int cnt;
int nump;
void init()
{
    memset(vis,0 ,sizeof(vis));
    memset(id,-1,sizeof(id));
    cnt = 0;
    nump = 2;
}
void add(int from,int to,double v)
{
    edge[cnt].to = to;
    edge[cnt].v = v;
    edge[cnt].pre = id[from];
    id[from] = cnt++;
}

double disget(int i,int j)
{
    return sqrt((ps[i].x - ps[j].x) * (ps[i].x - ps[j].x) +
                (ps[i].y - ps[j].y) * (ps[i].y - ps[j].y));
}
queue<int>q;
void spfa(int s)
{
    q.push(s);
    vis[s] = 1;
    d[s] = 0;
    while(q.size())
    {
        int now = q.front();q.pop();
        vis[now] = 0;
        for(int i = id[now];~i;i = edge[i].pre)
        {
            if(d[edge[i].to] > d[now] + edge[i].v)
            {
                d[edge[i].to] = d[now] + edge[i].v;
                if(!vis[edge[i].to])
                {
                    vis[edge[i].to] = 1;
                    q.push(edge[i].to);
                }
            }
        }
    }
}
int main()
{
    init();
    double x,y;
    scanf("%lf%lf",&x,&y);
    ps[0]=Point(x,y);
    scanf("%lf%lf",&x,&y);
    ps[1] = Point(x,y);
    d[1] = inf;
    while(~scanf("%lf%lf",&x,&y))
    {
        ps[nump] = Point(x,y);
        d[nump++] = inf;
        while(~scanf("%lf%lf",&x,&y) && x != -1 && y != -1)
        {
            ps[nump] = Point(x,y);
            double v = disget(nump,nump-1) / (1000.0 * 40.0) * 60.0;
            //cout<<v<<endl;
            add(nump - 1,nump,v);
            add(nump,nump - 1,v);
            d[nump++] = inf;
        }
    }
    for(int i = 0;i < nump;i++)
    {
        for(int j = 0;j < nump;j++)
        {
            double v = disget(i,j) / (1000 * 10) * 60;
            //cout<<v<<endl;
            add(i,j,v);
        }
    }
    spfa(0);
    printf("%d\n",int(d[1] + 0.5));
    return 0;
}