【序言】求單源最短路徑一直是個很熱門的話題。網上，大家也在爭相比試dijkstra和SPFA的優越性。遺憾的是，我一直沒學過dijkstra的堆優化，於是打算好好學習一下。以下是poj上隨便找的一道最短路的題目。

【dijkstra的堆優化】試想，在普通的dij中，每次尋找的當前dis的最小值時，要花費O（N）的效率，這樣實在太不佳了。我們基本的思想是構造一個最小堆，每次找最小值的時候直接取出堆頂即可。詳見這裡。

【STL的基本操作】

1.標頭檔案 #include<queue>

2.定義 priority_queue<x,vector<y>,greater<z

> >q;      -------->>>>>x、y、z是型別名，q是堆的變數名。其中vector是容器，greater可以使自己定義的bool型別(如sort的cmp)。當然，greater是系統已經定義好了的。

3.使用 q.push(x) 往堆中壓入x這個值 

           q.size() 查詢堆中的元素 

           q.pop() 彈出堆頂

           q.top() 返回堆頂元素的值

【注意】我又新定義了一種點對型別，定義方法：typedef pair<x,y>q;x和y是點對型別中每個元素的型別。這樣有什麼好處呢？因為dij的堆中我要存兩個元素：到此點的最小值和這個點的編號。用點對十分方便。

使用方法：make_pair(x,y)。返回點對的型別。

【題目】

Subway

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 5708	Accepted: 1846

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

Source

【大意】

給出起點、終點的座標。你要從起點走到終點。步行的速度是10km/h。（當然，你在兩個點之間走，肯定是勾股距離最短）。再給出N條地鐵線路，每條給出不小於2個的座標，表示這些座標構成一條線路。在地鐵軌道中，速度是40km/h。我們認為無論何時你到某個站，都恰好有一輛車來。求起點到終點的最小時間。

【程式碼】

#include<stdio.h>
#include<queue>
#include<cmath>
#include<cstring>
using namespace std;
const int maxn=200+5;
double d[maxn],map[maxn][maxn],ans;
int x[maxn],y[maxn],n,i;
bool flag[maxn];
typedef pair<double,int>Pair;
double dis(int x1,int y1,int x2,int y2)
{
  return sqrt(double((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)))/1000;
}
double dijkstra(int start,int end)
{
  for (int i=1;i<=n;i++) d[i]=2100000000;
  memset(flag,0,sizeof(flag));
  d[start]=0;
  priority_queue< Pair,vector<Pair>,greater<Pair> >q;
  q.push(make_pair(d[start],start));
  while (!q.empty())
  {
    Pair top=q.top();
    q.pop();
    int now=top.second;
    if (flag[now]) continue;
    flag[now]=true;
    for (int j=1;j<=n;j++)
      if ((!flag[j])&&(map[now][j]+d[now]<d[j]))
      {
        d[j]=d[now]+map[now][j];
        q.push(make_pair(d[j],j));
      }
  }
  return d[end];
}
void read()
{
  for (int i=1;i<maxn;i++)
    for (int j=1;j<maxn;j++)
      map[i][j]=2100000000;
  int start=n+1,xx,yy;
  while (scanf("%d%d",&xx,&yy)!=EOF)
  {
    if (xx==-1&&yy==-1)
    {
      for (int i=start;i<n;i++)
      {
        double temp=dis(x[i],y[i],x[i+1],y[i+1])/40;
        if (temp<map[i][i+1]) map[i][i+1]=map[i+1][i]=temp;
      }
      start=n+1;
      continue;
    }
    n++;
    x[n]=xx;y[n]=yy;
  }
  for (int i=1;i<n;i++)
    for (int j=i+1;j<=n;j++)
    {
      double temp=dis(x[i],y[i],x[j],y[j])/10;
      if (temp<map[i][j]) map[i][j]=map[j][i]=temp;
    }
}
  
int main()
{
  scanf("%d%d%d%d",&x[1],&y[1],&x[2],&y[2]);
  n=2;read();
  ans=dijkstra(1,2);ans*=60;
  printf("%.f",ans);
  return 0;
}