[poj2349]Arctic Network(最小生成树+贪心)

Arctic Network
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 17758   Accepted: 5646

Description

The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver and some outposts will in addition have a satellite channel. 
Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed D, which depends of the power of the transceivers. Higher power yields higher D but costs more. Due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts. 

Your job is to determine the minimum D required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.

Input

The first line of input contains N, the number of test cases. The first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S < P <= 500, the number of outposts. P lines follow, giving the (x,y) coordinates of each outpost in km (coordinates are integers between 0 and 10,000).

Output

For each case, output should consist of a single line giving the minimum D required to connect the network. Output should be specified to 2 decimal points.

Sample Input

1
2 4
0 100
0 300
0 600
150 750

Sample Output

212.13

Source

 
题意:
在给定坐标系中有p个点,可以建s条边权为零的边连接任意两点,问要使所有点联通且边权之和最小,建的边权最大的权值
(算了还是看别人翻译吧)
最小生成树裸题,贪心的去掉最小生成树中最大的s条边即可
 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 #include<math.h>
 5 #include<algorithm>
 6 using namespace std;
 7 typedef struct{
 8     int frm,to;
 9     double dis;
10 }edge;
11 typedef struct{
12     int x,y;
13 }point;
14 edge gra[250010];
15 point poi[510];
16 int fa[510],num;
17 int fnd(int x){
18     return fa[x]==x?x:fnd(fa[x]);
19 }
20 int uni(int x,int y){
21     int fx=fnd(x);
22     int fy=fnd(y);
23     fa[fy]=fx;
24     return 0;
25 }
26 int cmp(const edge &a,const edge &b){
27     return a.dis<b.dis;
28 }
29 int add(int frm,int to,double dis){
30     gra[++num].frm=frm;
31     gra[num].to=to;
32     gra[num].dis=dis;
33     return 0;
34 }
35 double kru(int k){
36     int cnt=0;
37     sort(gra+1,gra+num+1,cmp);
38     for(int i=1;i<=num;i++){
39         int fx=fnd(gra[i].frm);
40         int fy=fnd(gra[i].to);
41         if(fx!=fy){
42             cnt++;
43             if(cnt==k){
44                 return gra[i].dis;
45             }
46             uni(fx,fy);
47         }
48     }
49     return 0.0;
50 }
51 int main(){
52     int s,p,t;
53     scanf("%d",&t);
54     while(t--){
55         num=0;
56         scanf("%d %d",&s,&p);
57         if(s==0)s=1;
58         for(int i=1;i<=p;i++)scanf("%d %d",&poi[i].x,&poi[i].y);
59         for(int i=1;i<=p;i++)fa[i]=i;
60         for(int i=1;i<=p;i++){
61             for(int j=i+1;j<=p;j++){
62                 double dis=sqrt((double)((poi[i].x-poi[j].x)*(poi[i].x-poi[j].x)+(poi[i].y-poi[j].y)*(poi[i].y-poi[j].y)));
63                 add(i,j,dis);
64                 add(j,i,dis);
65             }
66         }
67         printf("%.2lf\n",kru(p-s));
68     }
69     return 0;
70 }
View Code

 

 
 
posted @ 2016-10-01 01:03  Pumbit-Legion  阅读(333)  评论(0编辑  收藏  举报