POJ2560 Freckles

 

Time Limit: 1000MS   Memory Limit: 65536KB   64bit IO Format: %lld & %llu

Description

In an episode of the Dick Van Dyke show, little Richie connects the freckles on his Dad's back to form a picture of the Liberty Bell. Alas, one of the freckles turns out to be a scar, so his Ripley's engagement falls through. 
Consider Dick's back to be a plane with freckles at various (x,y) locations. Your job is to tell Richie how to connect the dots so as to minimize the amount of ink used. Richie connects the dots by drawing straight lines between pairs, possibly lifting the pen between lines. When Richie is done there must be a sequence of connected lines from any freckle to any other freckle.

Input

The first line contains 0 < n <= 100, the number of freckles on Dick's back. For each freckle, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the freckle.

Output

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the freckles.

Sample Input

3
1.0 1.0
2.0 2.0
2.0 4.0

Sample Output

3.41

Source

Waterloo local 2000.09.23
 
最小生成树。注意double
 
 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 int read(){
 9     int x=0,f=1;char ch=getchar();
10     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
11     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
12     return x*f;
13 }
14 const int mxn=120;
15 int n;
16 struct point{
17     double x;double y;
18 }p[mxn];
19 double dist(point a,point b){
20     return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );
21 }
22 struct edge{
23     int x,y;
24     double dis;
25 }e[mxn*mxn];int cnt=0;
26 int cmp(edge a,edge b){
27     return a.dis<b.dis;
28 }
29 int fa[mxn];
30 void init(){
31     for(int i=1;i<=n;i++)fa[i]=i;
32 }
33 int find(int x){
34     if(fa[x]==x)return x;
35     return fa[x]=find(fa[x]);
36 }
37 double ans=0;
38 void Kruskal(){
39     int i,j;
40     int ti=0;
41     ans=0;
42     for(i=1;i<=cnt;i++){
43         int u=find(e[i].x),v=find(e[i].y);
44         if(u==v)continue;
45         ti++;
46         ans+=e[i].dis;
47         fa[u]=v;
48         if(ti==n-1)break;
49     }
50     return;
51 }
52 int main(){
53     int i,j;
54     while(scanf("%d",&n)!=EOF){
55         init();
56         cnt=0;
57         int u,v,dis;int m;
58         int i,j;
59         for(i=1;i<=n;i++){
60             scanf("%lf%lf",&p[i].x,&p[i].y);
61         }
62         for(i=1;i<n;i++)
63             for(j=i+1;j<=n;j++){
64                 e[++cnt]=(edge){i,j,dist(p[i],p[j])};
65             }
66         sort(e+1,e+cnt+1,cmp);
67         Kruskal();
68         printf("%.2f\n",ans);
69     }
70     return 0;
71 }

 

posted @ 2016-09-20 23:26  SilverNebula  阅读(244)  评论(0)  编辑  收藏  举报
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