USACO 2007 Dec Silver 2.Building Roads

做了 第一道 最小生成樹 hehe

Time Limit: 1 Sec Memory Limit: 64 MB
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
HINT
INPUT DETAILS:

Four farms at locations (1,1), (3,1), (2,3), and (4,3). Farms 1 and 4 are
connected by a road.

OUTPUT DETAILS:

Connect farms 1 and 2 with a road that is 2.00 units long, then connect
farms 3 and 4 with a road that is 2.00 units long. This is the best we can
do, and gives us a total of 4.00 unit lengths.

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<functional>
using namespace std;
typedef long long ll;
int n,m;
struct node
{
    int from,to;
    double val;
}edge[1000006];
int cnt;
int pos[1003][2];
bool cmp(node x,node y)
{
    return x.val<y.val;
}
int fa[1003];
int find(int x)
{
    if(fa[x]!=x) fa[x]=find(fa[x]);
    return fa[x];
}
void unnion(int x,int y)
{
    x=find(x),y=find(y);
    fa[x]=y;
}
int main()
{
    int i,j;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++) fa[i]=i;
    for(i=1;i<=n;i++)
    {
        scanf("%d%d",&pos[i][0],&pos[i][1]);
        for(j=1;j<i;j++)
        {
            edge[++cnt].from=i;
            edge[cnt].to=j;
            ll p=pos[i][0]-pos[j][0];
            ll q=pos[i][1]-pos[j][1];
            edge[cnt].val=sqrt(p*p+q*q);
        }
    }
    int kkk=0;
    for(i=1;i<=m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        unnion(x,y);
        kkk++;
    }
    sort(edge+1,edge+cnt+1,cmp);
    double sum=0.0;
    for(i=1;i<=cnt;i++)
    {
        if(kkk==n-1) break;
        if(find(edge[i].from)!=find(edge[i].to))
        {
            sum+=edge[i].val;
            kkk++;
            unnion(edge[i].from,edge[i].to);
        }
    }
    printf("%.2lf",sum);
    return 0;
}
/*

4 1
1000000 1000000
3000000 1000000
2000000 3000000
4000000 3000000
1 4

*/