POJ3625 Building Roads

Description

Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms.

Each of the N (1 ≤ N ≤ 1,000) farms (conveniently numbered 1..N) is represented by a position (X_i, Y_i) on the plane (0 ≤ X_i ≤ 1,000,000; 0 ≤ Y_i ≤ 1,000,000). Given the preexisting M roads (1 ≤ M ≤ 1,000) as pairs of connected farms, help Farmer John determine the smallest length of additional roads he must build to connect all his farms.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Two space-separated integers: X_i and Y_i 
* Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farm i and farm j.

Output

* Line 1: Smallest length of additional roads required to connect all farms, printed without rounding to two decimal places. Be sure to calculate distances as 64-bit floating point numbers.

Sample Input

4 1
1 1
3 1
2 3
4 3
1 4

Sample Output

4.00

Source

USACO 2007 December Silver

kruskal最小生成树

先处理出每两个点之间的距离，加入到边集。读入的已连接的点对也加入边集，距离为0。跑一遍kruskal出解

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=1020;
 9 int n,m;
10 struct point{
11     double x,y;
12 }p[mxn];
13 struct edge{
14     int x,y;
15     double dis;
16 }e[mxn*mxn];
17 int cmp(edge a,edge b){
18     return a.dis<b.dis;
19 }
20 double dist(point a,point b){
21     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
22 }
23 //
24 int fa[mxn];
25 int find(int x){
26     if(fa[x]==x)return x;
27     return fa[x]=find(fa[x]);
28 }
29 int cnt=0;
30 double ans=0;
31 void kruskal(){
32     int now=0;
33     for(int i=1;i<=cnt && now<n-1;i++){
34         int x=find(e[i].x);
35         int y=find(e[i].y);
36         if(x!=y){
37             ans+=e[i].dis;
38             fa[x]=y;now++;
39         }
40     }
41     printf("%.2f\n",ans);
42     return;
43 }
44 int main(){
45     scanf("%d%d",&n,&m);
46     int i,j;
47     for(i=1;i<=n;i++) fa[i]=i;
48     for(i=1;i<=n;i++)
49         scanf("%lf%lf",&p[i].x,&p[i].y);
50     for(i=1;i<n;i++)
51       for(j=i+1;j<=n;j++){
52           e[++cnt].dis=dist(p[i],p[j]);
53           e[cnt].x=i;e[cnt].y=j;
54       }
55     for(i=1;i<=m;i++){
56         cnt++;
57         scanf("%d%d",&e[cnt].x,&e[cnt].y);
58         e[cnt].dis=0;
59     }
60     sort(e+1,e+cnt+1,cmp);
61     kruskal();
62     return 0;
63 }