uva 10034 Freckles (kruskal||prim)

题目上只给的坐标，没有给出来边的长度，无论是prim算法还是kruskal算法我们都需要知道边的长度来操作。

这道题是浮点数，也没啥大的区别，处理一下就可以了。有关这两个算法的介绍前面我已经写过了，就不在多写了

prim算法：

<span style="font-family:Courier New;font-size:18px;">#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<algorithm>
#include<climits>
using namespace std;
struct node
{
	double i,j;
}g[105];
double gra[105][105];
double dist(double a,double b,double c,double d)
{
	return sqrt((a-c)*(a-c)+(b-d)*(b-d));
}
int n,cnt=0,T;
void prim()
{
	int visit[105],now,i,j;
	double dis[105];
	double Min;
	memset(visit,0,sizeof(visit));
	for(i=1; i<=n; i++)
		dis[i] = INT_MAX;
	visit[1] = 1, dis[1] = 0, now = 1;//now都是当前新加的点 
	for(i=1; i<=n; i++)
	{
		for(j=1; j<=n; j++)
		{
			if(!visit[j] && dis[j]>gra[now][j])//用新加的点来更新其他点到此集合的距离 
				dis[j] = gra[now][j];
		}
		Min = INT_MAX;
		for(j=1; j<=n; j++)
		{
			if(!visit[j] && dis[j] < Min)//每次都找到距离最小的点，加进去 
				Min = dis[now = j];
		}
		visit[now] = 1;
	} 
	double sum = 0;
	for(i=1; i<=n; i++)
	{
		sum += dis[i];
	}
	printf("%.2lf\n",sum);
	if(cnt!= T)//注意每两个输出案例之间都有一个换行 
		cout << endl;
}
int main()
{
	int i,j;
	cin >> T;
	while(cin >> n)
	{
		cnt ++;
		for(i=1; i<=n; i++)
		{
			cin >> g[i].i >> g[i].j;				
		} 
		memset(gra,0,sizeof(gra));
		for(i=1; i<=n; i++)
		{
			for(j=i+1; j<=n; j++)
			{
				gra[i][j] = gra[j][i] = dist(g[i].i,g[i].j,g[j].i,g[j].j);
			}
		}
		prim();
	}
	return 0;
}
</span>

kruskal算法:

<span style="font-family:Courier New;font-size:18px;">#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<algorithm>
#include<climits>
using namespace std;
struct node
{
	int i,j;
	double len;
}gra[10005];
struct node1
{
	double i,j;
}g[105];
int p[105];
double dist(double a,double b,double c,double d)
{
	return sqrt((a-c)*(a-c)+(b-d)*(b-d));
}
int cmp(const void *a,const void *b)
{
	return (((node *)a)->len - ((node *)b)->len > 0) ? 1:-1;
}
int n,cnt=0,T,k;
int find(int x)
{
	return x == p[x]? x: p[x] = find(p[x]);
}
void kruskal()
{
	double sum = 0;
	int i;
	for(i=1; i<k; i++)
	{
		int x = find(gra[i].i);
		int y = find(gra[i].j);
		if(x!=y)
		{
			sum += gra[i].len;
			p[x] = y;
		}
	}
	printf("%.2f\n",sum);
	if(cnt != T)
		cout << endl;
}
int main()
{
	int i,j;
	cin >> T;
	while(cin >> n)
	{
		cnt ++;
		for(i=1; i<=n; i++)
		{
			cin >> g[i].i >> g[i].j;				
		} 
		memset(gra,0,sizeof(gra));
		k=1;
		for(i=1; i<=n; i++)
		{
			for(j=i+1; j<=n; j++)
			{
				gra[k].len = dist(g[i].i,g[i].j,g[j].i,g[j].j);
				gra[k].i = i;
				gra[k].j = j;
				k++;
			}
		}
		//prim();
		for(i=1; i<=n; i++)
			p[i] = i;
		qsort(gra+1,k-1,sizeof(gra[0]),cmp);
		kruskal();
	}
	return 0;
}
</span>