POJ 2502 Subway / NBUT 1440 Subway / SCU 2186 Subway（图论，最短距离）
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
Http
POJ:https://vjudge.net/problem/POJ-2502
NBUT:https://vjudge.net/problem/NBUT-1440
SCU:https://vjudge.net/problem/SCU-2186
Source
图论，最短路径
题目大意
给出家和学校的坐标以及若干条地铁线及地铁站，并给出人走路和坐地铁的速度，求从家到学校的最短时间。
解决思路
算法还是比较好想，就是直接跑最短路就可以，但是有些细节比较麻烦。
首先，只有相邻的地铁站可以通过地铁相连，其他的点都要计算欧几里得距离。
另外，最后输出答案只要整数部分，这点题目中并没有说明。
代码
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<stack>
#include<cmath>
using namespace std;
const int maxN=300;
const int maxM=maxN*maxN;
const double v1=10000.0/60.0;//人的速度，均统一转换成m/min单位
const double v2=40000.0/60.0;//地铁的速度
const int inf=2147483647;
class Pos//坐标的结构体
{
public:
int x,y;
};
int n,m;
class Graph//图
{
private:
int cnt;
int Head[maxN];
int Next[maxM];
int V[maxM];
double W[maxM];
bool instack[maxN];
stack<int> S;//听说spfa+stack快于spfa+queue
public:
double Dist[maxN];//距离
void init()
{
cnt=0;
memset(Head,-1,sizeof(Head));
memset(Next,-1,sizeof(Next));
}
void Add_Edge(int u,int v,double w)
{
cnt++;
Next[cnt]=Head[u];
V[cnt]=v;
W[cnt]=w;
Head[u]=cnt;
}
void spfa(int s)//Spfa
{
memset(Dist,127,sizeof(Dist));
memset(instack,0,sizeof(instack));
Dist[s]=0;
instack[s]=1;
S.push(s);
do
{
int u=S.top();
S.pop();
instack[u]=0;
for (int i=Head[u];i!=-1;i=Next[i])
{
if (Dist[V[i]]>Dist[u]+W[i])
{
Dist[V[i]]=Dist[u]+W[i];
if (instack[V[i]]==0)
{
instack[V[i]]=1;
S.push(V[i]);
}
}
}
}
while (!S.empty());
}
void OutEdge()//为了方便检查输出的边
{
for (int i=1;i<=n;i++)
{
for (int j=Head[i];j!=-1;j=Next[j])
{
cout<<i<<"->"<<V[j]<<' '<<W[j]<<endl;
}
cout<<endl;
}
}
};
Pos P[maxN];
Graph G;
int read();
inline double Dist(Pos A,Pos B);
istream &operator >> (istream &is,Pos &p)//方便读入，重载一下输入运算符
{
is>>p.x>>p.y;
return is;
}
int main()
{
cin>>P[1]>>P[2];
G.init();
n=2;
while (cin>>P[n+1])//输入处理，有些麻烦
{
n++;
int now=n;
Pos input;
for (int i=1;i<=n;i++)
{
double dist=Dist(P[n],P[i])/v1;
G.Add_Edge(n,i,dist);
G.Add_Edge(i,n,dist);
}
while (cin>>input)
{
if (input.x==-1)
break;
n++;
P[n]=input;
double dist=Dist(P[n],P[n-1])/v2;
G.Add_Edge(n,n-1,dist);
G.Add_Edge(n-1,n,dist);
for (int i=1;i<=n;i++)
{
//cout<<"Link:"<<i<<' '<<n<<endl;
double dist2=Dist(P[n],P[i])/v1;
G.Add_Edge(n,i,dist2);
G.Add_Edge(i,n,dist2);
}
}
}
//G.OutEdge();
G.spfa(1);
printf("%.0f
",G.Dist[2]);
return 0;
}
int read()
{
int x=0;
int k=1;
char ch=getchar();
while (((ch>'9')||(ch<'0'))&&(ch!='-'))
ch=getchar();
if (ch=='-')
{
k=-1;
ch=getchar();
}
while ((ch>='0')&&(ch<='9'))
{
x=x*10+ch-48;
ch=getchar();
}
return x*k;
}
inline double Dist(Pos A,Pos B)
{
return sqrt((double)((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)));
}
按照我这个方法建出来的图会有部分边重复，但没有关系。