poj 3625 Building Roads (prim)

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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_iand 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

Sample Output

4.00

#include<stdio.h>
#include<string.h>
#include<cmath>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int pre[10003];
int vis[10003];
double dis[10003];
double x[1003],y[1003];
double Side[1003][1003];
int n,m;
double res=0;
double Distance(double x1,double y1,double x2,double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
void prime()
{
    res=0;
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
        dis[i]=Side[1][i];
    dis[1]=0;
    vis[1]=1;
    int k;
    double Min=inf;
    for(int i=1;i<n;i++)
    {
        Min=inf;
        for(int j=1;j<=n;j++)
        {
            if(dis[j]<Min&&!vis[j])
                Min=dis[j],k=j;
        }
        //printf("%.2f\n",Min);
        res+=Min;
        vis[k]=1;
        for(int j=1;j<=n;j++)
        {
            if(dis[j]>Side[k][j]&&!vis[j])
                dis[j]=Side[k][j];
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",&x[i],&y[i]);
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
            Side[i][j]=inf;
        for(int i=1;i<=n;i++)Side[i][i]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                Side[i][j]=Side[j][i]=Distance(x[i],y[i],x[j],y[j]);
            }
        }
        int a,b;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            Side[a][b]=Side[b][a]=0;
        }
        prime();
        printf("%.2f\n",res);
    }
}