Object Clustering
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 dij = |ai - aj| |bi - bj|, and then we say i is dij 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 dij ≤ 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
6 2 1 2 2 3 2 2 3 4 4 3 3 1 Sample Output
2 Source
POJ Monthly–2007.08.05, Li, Haoyuan 题解
代码语言：javascript复制#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<bitset>
#include<set>
#include<stack>
#include<map>
#include<list>
#include<new>
#include<vector>
#define MT(a,b) memset(a,b,sizeof(a));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double pi=acos(-1.0);
const double E=2.718281828459;
const ll mod=1e8 7;
const int INF=0x3f3f3f3f;
int n,k;
struct node{
int x;
int y;
int id;
bool friend operator<(node a,node b){
return a.x==b.x?a.y<b.y:a.x<b.x;
///保证树状数组更新和查询时不会遗漏
}
}point[10005];
struct edge{
int s;
int e;
int c;
bool friend operator<(edge a,edge b){
return a.c<b.c;
}
}load[40000];
int sign=0;
int p[10005];
int find(int x){
return p[x]==x?x:p[x]=find(p[x]);
}
void kruskal(){
for(int i=1;i<=n;i )
p[i]=i;
sort(load 1,load 1 sign);
int cnt=0;
for(int i=1;i<=sign;i ){
int x=find(load[i].s);
int y=find(load[i].e);
if(x!=y){
cnt ;
p[x]=y;
if(cnt==n-k){
printf("%dn",load[i].c);
return ;
}
}
}
}
int id[10005]; ///y-x为索引的编号
int xy[10005]; ///y-x为索引 x y的最小值
void update(int index,int minn,int s) ///index:y-x minn:x y s:编号
{
index =1000;
for(int i=index;i>=1;i-=(i&(-i))){
if(xy[i]>minn){
xy[i]=minn;
id[i]=s;
}
}
}
void query(int index,int minn,int s) ///index:y-x minn:x y s:编号
{
index =1000;
int e=-1,c=INF;
///现在以编号s为原点,查询y-x>=index的点中x y的最小值
for(int i=index;i<10000;i =(i&(-i))){
if(xy[i]<c){
e=id[i];
c=xy[i];
}
}
if(e!=-1)
load[ sign]=edge{s,e,c-minn};
}
void build_edge()
{
/// 以(xi,yi)为原点,对于第1区域内的点(x,y)满足条件
/// x>=xi,y-x>=yi-xi,(x y)最小
sort(point 1,point 1 n);
memset(id,-1,sizeof(id));
fill(xy,xy 10005,INF);
///按照x升序
///保证后面查询时,x都比当前的x大
for(int i=n;i>=1;i--){
int index=point[i].y-point[i].x;
int minn=point[i].x point[i].y;
query(index,minn,point[i].id);
update(index,minn,point[i].id);
}
}
int main() ///第K大边
{
scanf("%d %d",&n,&k);
for(int i=1;i<=n;i ){
scanf("%d %d",&point[i].x,&point[i].y);
point[i].id=i;
}
///1象限建边
build_edge();
///2象限建边
for(int i=1;i<=n;i )
swap(point[i].x,point[i].y);
build_edge();
///3象限建边
for(int i=1;i<=n;i )
point[i].x=-point[i].x;
build_edge();
///4象限建边
for(int i=1;i<=n;i )
swap(point[i].x,point[i].y);
build_edge();
kruskal();
return 0;
}
