Poj--2502（图论，最短路，建图）

2014-09-10 22:53:55

思路：裸的Dij，最短路，建图的时候有点trick，地铁只能一站一站地开，还有就是各个点的坐标最好用double存，存int可能会出问题。

 1 /*************************************************************************
 2     > File Name: 2502.cpp
 3     > Author: Nature
 4     > Mail: 564374850@qq.com 
 5     > Created Time: Wed 10 Sep 2014 08:55:55 PM CST
 6 ************************************************************************/
 7 
 8 #include <cstdio>
 9 #include <cstring>
10 #include <cstdlib>
11 #include <cmath>
12 #include <vector>
13 #include <queue>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17 typedef long long ll;
18 const int INF = 1 << 30;
19 
20 int cnt;
21 double x[205],y[205];
22 double g[205][205];
23 double d[205];
24 
25 double Dist(double x1,double y1,double x2,double y2){
26     return sqrt((x2 - x1)*(x2 - x1) + 1.0*(y2 - y1)*(y2 - y1));
27 }
28 
29 void Dij(){
30     int vis[205];
31     memset(vis,0,sizeof(vis));
32     for(int i = 0; i < cnt; ++i) d[i] = INF;
33     d[0] = 0;
34     for(int i = 0; i < cnt; ++i){
35         int p;
36         double tmin = (double)INF;
37         for(int j = 0; j < cnt; ++j) if(!vis[j] && d[j] < tmin)
38             tmin = d[p = j];
39         vis[p] = 1;
40         for(int j = 0; j < cnt; ++j) if(d[j] > d[p] + g[p][j])
41             d[j] = d[p] + g[p][j];
42     }
43 }
44 
45 void Init(){
46     for(int i = 0; i < 205; ++i){
47         for(int j = 0; j < 205; ++j){
48             g[i][j] = INF;
49         }
50         g[i][i] = 0;
51     }
52 }
53 
54 int main(){
55     int x1,y1,x2,y2;
56     Init();
57     cnt = 2;
58     scanf("%lf%lf%lf%lf",&x[0],&y[0],&x[1],&y[1]);
59     while(scanf("%lf%lf",&x[cnt],&y[cnt]) != EOF){
60         ++cnt;
61         while(scanf("%lf%lf",&x[cnt],&y[cnt]) != EOF){
62             if(x[cnt] == -1 && y[cnt] == -1) break;
63             ++cnt;
64             g[cnt-2][cnt-1] = g[cnt-1][cnt-2] 
65                 = Dist(x[cnt-2],y[cnt-2],x[cnt-1],y[cnt-1]) / 40000.0;
66         }
67     }
68     for(int i = 0; i < cnt; ++i)
69     for(int j = i + 1; j < cnt; ++j)
70         g[i][j] = g[j][i] = min(g[i][j],Dist(x[i],y[i],x[j],y[j]) / 10000.0);
71     Dij();
72     printf("%.0lf\n",d[1] * 60.0);
73     return 0;
74 }

 

posted @ 2014-09-10 22:55  Naturain  阅读(162)  评论(0)    收藏  举报