[poj2349]Arctic Network(最小生成樹+貪心) - Pumbit-Legion @ IT工程師數位筆記本

文章出處

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

Sample Output

212.13

Source

Waterloo local 2002.09.28

題意：

在給定坐標系中有p個點，可以建s條邊權為零的邊連接任意兩點，問要使所有點聯通且邊權之和最小，建的邊權最大的權值

(算了還是看別人翻譯吧)

最小生成樹裸題，貪心的去掉最小生成樹中最大的s條邊即可

 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 }