poj 3625 Building Roads

poj 3625 Building Roads

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

題意：n個村莊，m條已經修好的道路，接下來n行表示n個村莊的坐標（xi，yi），m行表示兩個村莊（a，b）已經修好道路。求n個村莊互相連通需要修建的最短道路。

思路：最小生成樹

Prim算法：

#include
#include
#include
#include
#define INF 999999
#define M 1005
using namespace std;

double map[M][M],len[M];   // 注意類型
int vis[M];
int x[M],y[M];

double dis(int i,int j)  // 任意兩個村莊的距離
{
	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>a>>b;
				map[a][b]=map[b][a]=0; // 巧妙處理。若已經修好道路，則長度賦值為0.
			}

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

			for(i=1;imap[temp][j])
						len[j]=map[temp][j];
				}
			}
			printf("%.2lf\n",sum);
	}
	return 0;
}