| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10803 | Accepted: 3062 |
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 (Xi, Yi) on the plane (0 ≤ Xi ≤ 1,000,000; 0 ≤ Yi ≤ 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: Xi and Yi
* 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
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 }
原文：http://www.cnblogs.com/SilverNebula/p/5872558.html