Building Roads

Time Limit: 1000MS

 

Memory Limit: 65536K

Total Submissions: 13247

 

Accepted: 3661

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

​​USACO 2007 December Silver​​

 

题意：

n个点的坐标,m已经建成的路,问你为使该图连通,最少还需要连接总长度为多少的边?

分析：

  算出任意两点间的距离,得到一个完全图,求最小生成树即可.已经连接上的边长度为0.用kruskal算法.

 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 1000+5;
const int dx[] = {-1,1,0,0,-1,-1,1,1};
const int dy[] = {0,0,-1,1,-1,1,-1,1};
using namespace std;
struct Node{
    double x,y;
}node[N];
struct Edge{
    int x,y;
    double dis;
    Edge(){}
    Edge(int x,int y,double dis):x(x),y(y),dis(dis){}
    
}edge[N*1000];
bool cmp(Edge a,Edge b)
{
     if(a.dis==b.dis)
        return a.x<b.x;
     else
        return a.dis<b.dis;
}
 int n,m;
int tot;
int father[N];
double G[N][N];
double getDis(int i,int j){
    return sqrt( fabs((node[i].x-node[j].x))*fabs((node[i].x-node[j].x)) + fabs((node[i].y-node[j].y))*fabs((node[i].y-node[j].y)) );
}
int Find(int x){
    if(father[x]==x)
        return x;
    return father[x]=Find(father[x]);
}

double mst;
int cnt=0;
void Kruskal(){
    sort(edge+1,edge+tot+1,cmp);

    for(int i=1;i<=tot;++i){
        int x=Find(edge[i].x);
        int y=Find(edge[i].y);
        if(x!=y){
            father[x]=y;
            mst+=edge[i].dis;
            cnt++;
            if(cnt==n-1)
                break;
        }
    }
}
int main(){
   
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        father[i]=i;
        
    for(int i=1;i<=n;i++)
        scanf("%lf%lf",&node[i].x,&node[i].y);

    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++){
    
            double temp=getDis(i,j);
            edge[++tot].x=i;
            edge[tot].y=j;
            edge[tot].dis=temp;
        }
    }
    for(int i=1;i<=m;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        edge[++tot].x=x;
        edge[tot].y=y;
        edge[tot].dis=0;
    }

    Kruskal();
    
    printf("%.2f",mst);
    return 0;
}