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poj 3625 Building Roads

2024-07-13 07:37:30   218人阅读

Building Roads
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8999 Accepted: 2596

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


题意：n个村庄，m条已经修好的道路，接下来n行表示n个村庄的坐标（xi，yi），m行表示两个村庄（a，b）已经修好道路。求n个村庄互相连通需要修建的最短道路。

思路：最小生成树

Prim算法：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#define INF 999999
#define M 1005
using namespace std;

double map[M][M],len[M];   // 注意类型
int vis[M];
int x[M],y[M];

double dis(int i,int j)  // 任意两个村庄的距离
{
	double k1,k2,k;
	k1=x[i]-x[j];
	k2=y[i]-y[j];
	k=sqrt(k1*k1+k2*k2);
	return k;
}

int main ()
{
	int n,m,a,b;
	int i,j;
	int temp;
	double sum;
	while(cin>>n>>m)
	{
		memset(map,0,sizeof(map));
		memset(vis,0,sizeof(vis));
		
		for(i=1;i<=n;i++)
			cin>>x[i]>>y[i];
		
		for(i=1;i<=n;i++)
			for(j=1;j<n;j++)
				map[i][j]=map[j][i]=dis(i,j);
			
			for(i=0;i<m;i++)
			{
				cin>>a>>b;
				map[a][b]=map[b][a]=0; // 巧妙处理。若已经修好道路，则长度赋值为0.
			}

			
			for(i=1;i<=n;i++)
				len[i]=map[1][i];
			
			vis[1]=1;
			sum=0.0;

			for(i=1;i<n;i++)
			{ 
				double min=INF;
				for(j=1;j<=n;j++)
				{
					if(!vis[j] && len[j]<min)
					{
						min=len[j];
						temp=j;	
					}
				}
				sum+=min;
				vis[temp]=1;
				
				for(j=1;j<=n;j++)
				{
					if(!vis[j] && len[j]>map[temp][j])
						len[j]=map[temp][j];
				}
			}
			printf("%.2lf\n",sum);
	}
	return 0;
}


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