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
题意：在xy轴第一区域上，一个人要从家去学校，途中他可以步行，也可以坐地铁，步行的速度是10km/h，地铁的速度是40km/h,坐标轴的单位距离是米，问他从家到学校最短要用
多少分钟时间。最多有200个地铁站点。输入第一行为四个数，是家和学校的坐标，然后往下每行为一条地铁线的坐标，每条地铁线的站点以-1，-1为输入结束。
思路：将每个站点比喻成一个点，点之间用时间连接，不能连接的点用无穷大表示，家和学校作为第一第二个点，其他地铁站点从三开始标记，最后用Dijkstra算法算出1到2之间的最短
时间按即可。注意结果为四舍五入的整数。
代码：
1 #include <cstdio>
2 #include <fstream>
3 #include <algorithm>
4 #include <cmath>
5 #include <deque>
6 #include <vector>
7 #include <queue>
8 #include <string>
9 #include <cstring>
10 #include <map>
11 #include <stack>
12 #include <set>
13 #include <sstream>
14 #include <iostream>
15 #define mod 998244353
16 #define eps 1e-6
17 #define ll long long
18 #define INF 0x3f3f3f3f
19 using namespace std;
20
21 //结构体存放点的坐标
22 struct node
23 {
24 int x,y;
25 };
26 //存放点的信息
27 node no[3000];
28 //点之间的信息
29 double ma[310][310];
30 //ans代表站点的标记
31 int ans;
32 //各点到起点之间的距离
33 double dis[210];
34 //是否为最短点标记
35 int vis[210];
36 void dijkstra()
37 {
38 //初始dis数组
39 for(int i=1;i<=ans;i++)
40 {
41 dis[i]=ma[1][i];
42 }
43 //初始化
44 memset(vis,0,sizeof(vis));
45 vis[1]=1;
46 for(int i=1;i<ans;i++)
47 {
48 double mi=INF;
49 int k;
50 //寻找最短路
51 for(int j=1;j<=ans;j++)
52 {
53 if(dis[j]<mi&&!vis[j])
54 {
55 mi=dis[j];
56 k=j;
57 }
58 }
59 //标记
60 vis[k]=1;
61 for(int j=1;j<=ans;j++)
62 {
63 //判别是否为最短路
64 if(!vis[j]&&dis[j]>dis[k]+ma[k][j])
65 {
66 dis[j]=dis[k]+ma[k][j];
67 }
68 }
69 }
70 }
71 //获取两点之间的距离
72 double len(node m,node n)
73 {
74 return sqrt(double((m.x-n.x)*(m.x-n.x)+(m.y-n.y)*(m.y-n.y)));
75 }
76 int main()
77 {
78 //初始化ma函数两点之间距离最大
79 for(int i=0;i<210;i++)
80 {
81 for(int j=0;j<210;j++)
82 {
83 if(i==j)
84 {
85 ma[i][j]=0;
86 }
87 else
88 {
89 ma[i][j]=ma[j][i]=INF;
90 }
91
92 }
93 }
94 //输入第一个点和第二个点
95 scanf("%d %d %d %d",&no[1].x,&no[1].y,&no[2].x,&no[2].y);
96 int a,b,bj;
97 //从第三个点开始存放地铁的信息；
98 //bj为当前地铁的初始站点的信息
99 bj=ans=3;
100 while(cin>>a>>b)
101 {
102 //如果为-1，则代表地铁站点输入玩了，接着存放下一个地铁站点
103 if(a==-1&&b==-1)
104 {
105 bj=ans;
106 continue;
107 }
108 //存放信息
109 no[ans].x=a;
110 no[ans].y=b;
111 //如果bj不等于ans表示当前ans站点与第bj站点是一个地铁线的；
112 if(bj!=ans)
113 {
114 //存放地铁线之间的距离
115 //以地铁速度存放
116 ma[ans][ans-1]=ma[ans-1][ans]=len(no[ans],no[ans-1])*3/2000;
117 }
118 ans++;
119 }
120 //遍历存放素有站点之间的关系
121 //以步行速度存放
122 for(int i=1;i<=ans;i++)
123 {
124 for(int j=i+1;j<=ans;j++)
125 {
126 ma[i][j]=ma[j][i]=min(ma[i][j],len(no[i],no[j])*3/500);
127 }
128 }
129 //求最短路
130 dijkstra();
131 //对结果四舍五入
132 int s=dis[2]+0.5;
133 printf("%d\n",s);
134 }