【POJ 2502】Subway（最短路dij）

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

题目大意就是说一个人走路的时速是10km/h，坐动车的话是40km/h，现在问你从（x,y）到达（posx，posy）需要花费的最小时间（单位为分钟）。每当输入-1 -1时表示一条线路的所有站点输入结束。

思路

这题主要难在输入的处理，需要计算两点间的距离，同时又要注意只有处在同一条线路上的相邻动车站之间才能到达，在建边的时候要特别注意这一点。对于多组输入可以用while来实现，单位可以在计算前先统一转换。这些处理好后就是一个dij的模版题了。因为对于这种卡输入的题我表示并不想做，所以就网上copy了一份代码，大家可以参考。

代码

#include<stdio.h>#include<math.h>#include<string.h>#define min(a,b) (a<b?a:b)#define INF 0x3f3f3f3f#define MAXV 205typedef struct{    int x,y;}Point;double map[MAXV][MAXV],d[MAXV];int v[MAXV],n;Point point[MAXV],a;int exist(Point t);float distance(Point a,Point b);void Dijkstra();int main(){    int subway[MAXV],k;//subway表示地铁上的第n个点    for(int i=1;i<=MAXV;i++)        for(int j=1;j<=MAXV;j++)            map[i][j]=(i==j?0:INF);    scanf("%d%d%d%d",&point[1].x,&point[1].y,&point[2].x,&point[2].y);    n=2;    while(~scanf("%d%d",&a.x,&a.y))    {        k=0;        subway[k++]=exist(a);        while(scanf("%d%d",&a.x,&a.y)&&a.x!=-1&&a.y!=-1)            subway[k++]=exist(a);        for(int i=1;i<k;i++)            map[subway[i]][subway[i-1]]=map[subway[i-1]][subway[i]]=distance(point[subway[i]],point[subway[i-1]])*60.0/40000.0;//40 km/h= 40000/60 m/min    }    for(int i=1;i<=n;i++)        for(int j=1;j<=i;j++)            map[i][j]=map[j][i]=min(map[i][j],distance(point[i],point[j])*60.0/10000.0);//10 km/h= 10000/60 m/min    Dijkstra();    return 0;}int exist(Point t)//返回值为第几个点 {    int i;    for(i=1;i<=n;i++)        if(t.x==point[i].x&&t.y==point[i].y)             break;    if(i==n+1)        point[++n]=t;    return i;}float distance(Point a,Point 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++)        d[i]=map[1][i];    memset(v,0,sizeof(v));    for(int i=1;i<=n;i++)    {        int x,m=INF;        for(int y=1;y<=n;y++)            if(!v[y]&&d[y]<m)                m=d[x=y];        v[x]=1;        for(int y=1;y<=n;y++)            d[y]=min(d[y],d[x]+map[x][y]);    }    printf("%d\n",int(d[2]+0.5));}