Building Roads

Building Roads

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 6364
Accepted: 1831

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 andY_i
* Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farmi 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 11 13 12 34 31 4

Sample Output

4.00

Source

USACO 2007 December Silver

#include<iostream>#include<cstdio>#include<cmath>using namespace std;double map[1001][1001];int N,M;double  prim(){ bool vis[1001]={0}; double lw[1001],min,sum; int k; for(int i=1;i<=N;i++)  lw[i]=map[1][i]; vis[1]=1; sum=0; for(int i=1;i<N;i++) {  min=10000001;  for(int j=1;j<=N;j++)   if(!vis[j] &&  lw[j]<min)   {    min=lw[j];    k=j;   }            sum+=min;   vis[k]=1;   for(int j=1;j<=N;j++)    if( !vis[j] && map[k][j]<lw[j])     lw[j]=map[k][j]; } return sum;}double dis(double x1,double y1,double x2,double y2){ return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}int main(){     double vx[1001],vy[1001],total;  int n,m;  while(scanf("%d%d",&N,&M) !=EOF)  {   for(int i=1;i<=N;i++)    scanf("%lf%lf",&vx[i],&vy[i]);   for(int i=1;i<=N;i++)    for(int j=i+1;j<=N;j++)    {     double temp=dis(vx[i],vy[i],vx[j],vy[j]);     map[i][j]=map[j][i]=temp;    }    for(int i=1;i<=M;i++)    {     scanf("%d%d",&n,&m);     map[n][m]=map[m][n]=0;    }    total=prim();    printf("%.2lf\n",total);  }}