Eddy's picture + prime算法 + kruskal算法

Eddy's picture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5029    Accepted Submission(s): 2482

Problem Description

Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested in his picture.Eddy feels very puzzled,in order to change all friends 's view to his technical of painting pictures ,so Eddy creates a problem for the his friends of you.
Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. How many distants does your duty discover the shortest length which the ink draws?

 

Input

The first line contains 0 < n <= 100, the number of point. For each point, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the point.

Input contains multiple test cases. Process to the end of file.

 

Output

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points.

 

Sample Input

3
1.0 1.0
2.0 2.0
2.0 4.0

 

Sample Output

3.41
 
/*思路：prime算法*/
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std ;
#define MAX 102 
#define inf 1000000000.0
typedef struct infor 
{
	double x , y ;
} infor ;
int n , vi[MAX] ;
infor a[MAX] ;
double s[MAX][MAX] ;
void prime()
{
	int i, j , k ;
	double sum = 0 , d[MAX] ;
	for( i = 0 ; i < n ; i ++)
		d[i] = s[0][i] ;
	d[0] = 0 ;
	vi[0] = 1;
	for( i = 1; i < n ; i ++)
	{
		double min1 = inf ;
		for( j = 0 ; j < n ; j ++)
			if(!vi[j] && min1 > d[j])
			{
				k = j ;
				min1 = d[j] ;
			}
		sum +=min1 ;
		if(sum >= inf )
			break;
		vi[k] = 1;
		for( j = 0 ; j < n ; j ++)
			if(!vi[j] && d[j] > s[k][j])
				d[j] = s[k][j] ;
	}
	cout <<fixed<<setprecision(2)<< sum << endl ;
}
int main(void)
{
	
	while( cin >> n )
	{
		memset(vi, 0 , sizeof(vi));
		int i , j ; 
		for( i = 0 ; i < n ; i ++ )
			cin >> a[i].x >> a[i].y ;
		for( i = 0 ; i < n ; i ++)
			for( j = 0 ; j < n ; j ++)
			{
				s[i][j] = s[j][i] = inf ; //初始化每两点的距离为无穷大
				double dis = sqrt((a[i].x - a[j].x)*(a[i].x - a[j].x) + (a[i].y - a[j].y )* (a[i].y - a[j].y ));  //the distance between two points
				s[i][j] = s[j][i] = s[i][j] > dis ? dis : s[i][j] ; //aviod the same point,take the smallest one 
			}
			prime(); //begin prime algorithm 
	}
	return 0;
}
 
/*思路：kruskal算法*/
#include<iostream>
#include<algorithm>
#include<cmath>
#include<iomanip>
using namespace std ;
#define MAX 102
int father[MAX];
typedef struct infor
{
	int start , end ;
	double dis ;
}infor ;
infor s[MAX*MAX] ;
int n ;
bool cmp(infor a , infor b)
{
	if(a.dis < b.dis )
		return true ;
	else 
		return false ;
}
int findset(int x)
{
	return x!=father[x] ? father[x] = findset(father[x]): x;
}
double kruskal(int m)
{
	int i  , x , y;
	double mincost= 0 ;
	for( i = 0 ; i < m ; i ++)
	{
		x= s[i].start ;
		y = s[i].end ;
		x= findset(x);
		y = findset(y);
		if(x!=y)
		{
			father[x] = y ;
			mincost +=s[i].dis ;
		}
	}
	return mincost ;
}
int main(void)
{
	double mincost = 0 ;
	while(cin >> n )
	{
		int i , j ;
		double a[MAX] ,b[MAX];
		for( i = 0 ; i < n ; i ++)
			cin >> a[i] >> b[i] ;
		int num = 0 ;
		for( i = 0 ; i < n ; i ++)
			for( j = 0 ; j < n ; j ++)
			{
				double dis = sqrt((a[i] - a[j])*(a[i] - a[j]) + (b[i] - b[j]) * (b[i] - b[j] ));
				s[num].start = i ;
				s[num].end = j ;
				s[num++].dis = dis ;
			}
		for( i = 0 ; i < n ; i ++)
			father[i] = i ;
		sort(s, s + num , cmp);
		mincost = kruskal(num) ;
		cout<<fixed<<setprecision(2)<<mincost<<endl;
	
	}
	return 0;
}