POJ 3241 Object Clustering-优快云博客

本文介绍了一种求解曼哈顿最小生成树问题的方法，该问题要求将N个对象根据其相似度指标分组，并找到使得各组内任意两对象间最大相似度最小的阈值X。通过一种特殊算法，利用树状数组记录每个方向上的最小距离并进行更新，最终求得最小生成树中第n-k条边的长度。

Description

We have N (N ≤ 10000) objects, and wish to classify them into several groups by judgement of their resemblance. To simply the model, each object has 2 indexes a and b (a, b ≤ 500). The resemblance of object i and object j is defined by d_ij = |a_i- a_j| + |b_i-b_j|, and then we say i is d_ij resemble to j. Now we want to find the minimum value of X, so that we can classify the N objects into K (K < N) groups, and in each group, one object is at most X resemble to another object in the same group, i.e, for every object i, if i is not the only member of the group, then there exists one object j (i ≠ j) in the same group that satisfies d_ij ≤ X

Input

The first line contains two integers N and K. The following N lines each contain two integers a and b, which describe a object.

Output

A single line contains the minimum X.

Sample Input

Sample Output

Source

POJ Monthly--2007.08.05, Li, Haoyuan

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曼哈顿最小生成树~

题目大意是求曼哈顿最小生成树中第n-k条边的长度，用来学曼哈顿最小生成树的~是莫队的基础？

求n个点曼哈顿距离最小生成树，如果用kruskal的话一一枚举建边会很慢，所以用到了这种神奇的算法。

对于每一个点，可能连边的点每45度都只有一个，所以用树状数组记录最小距离，然后每个方向更新~神奇的算法~

#include<cstdio>
#include<cstring>
#include<climits>
#include<iostream>
#include<algorithm>
using namespace std;

int n,k,fa[10001],c[4001],d[4001],tot;

struct node{
	int x,y,num;
}a[10001];

struct node1{
	int x,y,v;
}ed[40001];

bool cmp1(node u,node v)
{
	return u.x==v.x ? u.y<v.y:u.x<v.x;
}

bool cmp2(node1 u,node1 v)
{
	return u.v<v.v;
}

int findd(int u)
{
	return fa[u]==u ? u:fa[u]=findd(fa[u]);
}

int lowbit(int u)
{
	return u&(-u);
} 

void query(int num,int pos,int val)
{
	pos+=1000;
	int now=-1,val1=INT_MAX;
	for(int i=pos;i<4000;i+=lowbit(i))
	  if(val1>c[i])
	  {
	  	val1=c[i];now=d[i];
	  }
	if(now!=-1)
	{
		ed[++tot].x=num;ed[tot].y=now;ed[tot].v=val1-val;
	}
}

void change(int num,int pos,int val)
{
	pos+=1000;
	for(int i=pos;i;i-=lowbit(i))
	  if(c[i]>val)
	  {
	  	c[i]=val;d[i]=num;
	  }
}

void deal()
{
	memset(c,63,sizeof(c));
	sort(a+1,a+n+1,cmp1);
	for(int i=n;i>=1;i--)
	{
		int pos=a[i].y-a[i].x,val=a[i].x+a[i].y;
		query(a[i].num,pos,val);
		change(a[i].num,pos,val);
	}
}

int ans()
{
	if(n==k) return 0;
	deal();
	for(int i=1;i<=n;i++) swap(a[i].x,a[i].y);deal();
	for(int i=1;i<=n;i++) a[i].y=-a[i].y;deal();
	for(int i=1;i<=n;i++) swap(a[i].x,a[i].y);deal();
	sort(ed+1,ed+tot+1,cmp2);
	for(int i=1;i<=tot;i++)
	  if(findd(ed[i].x)!=findd(ed[i].y))
	  {
	  	fa[findd(ed[i].x)]=findd(ed[i].y);
	  	n--;if(n==k) return ed[i].v;
	  }
}

int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y),a[i].num=i;
	printf("%d\n",ans());
	return 0; 
}