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 dijX

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

又是一道曼哈顿距离最小生成树。。。
code:
 #include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#define maxn 100005
#define inf 1061109567
using namespace std;
char ch;
bool ok;
void read(int &x){
for (ok=,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=;
for (x=;isdigit(ch);x=x*+ch-'',ch=getchar());
if (ok) x=-x;
}
struct Point{
int x,y,d,id;
}point[maxn],tmp[maxn];
bool cmp1(Point a,Point b){
if (a.x!=b.x) return a.x>b.x;
return a.y>b.y;
}
int calc(Point a,Point b){return abs(a.x-b.x)+abs(a.y-b.y);}
int n,m,k,ans;
struct DATA{
int val,pos;
void init(){val=inf,pos=-;}
void update(DATA b){if (val>b.val) val=b.val,pos=b.pos;}
};
int d[maxn],cntd;
struct bit{
#define lowbit(x) ((x)&(-(x)))
DATA node[maxn];
void init(){for (int i=;i<=cntd;i++) node[i].init();}
void insert(int x,DATA p){for (int i=cntd-x+;i<=cntd;i+=lowbit(i)) node[i].update(p);}
int query(int x){
DATA ans; ans.init();
for (int i=cntd-x+;i;i-=lowbit(i)) ans.update(node[i]);
return ans.pos;
}
}T;
struct Edge{
int u,v,c;
}edge[maxn<<];
bool cmp2(Edge a,Edge b){return a.c<b.c;}
void prepare(){
for (int i=;i<=n;i++) d[i]=point[i].d=point[i].y-point[i].x;
sort(d+,d+n+),cntd=unique(d+,d+n+)-d-;
for (int i=;i<=n;i++) point[i].d=lower_bound(d+,d+cntd+,point[i].d)-d;
sort(point+,point+n+,cmp1),T.init();
for (int i=;i<=n;i++){
int u=point[i].id,v=T.query(point[i].d);
if (v!=-) edge[++m]=(Edge){u,v,calc(tmp[u],tmp[v])};
T.insert(point[i].d,(DATA){point[i].x+point[i].y,u});
}
}
int fa[maxn];
int find(int x){return x==fa[x]?fa[x]:fa[x]=find(fa[x]);}
int main(){
read(n),read(k);
for (int i=;i<=n;i++) read(point[i].x),read(point[i].y),point[i].id=i;
for (int i=;i<=n;i++) tmp[i]=point[i]; prepare();
for (int i=;i<=n;i++) point[i].x=tmp[i].y,point[i].y=tmp[i].x,point[i].id=i; prepare();
for (int i=;i<=n;i++) point[i].x=-tmp[i].y,point[i].y=tmp[i].x,point[i].id=i; prepare();
for (int i=;i<=n;i++) point[i].x=tmp[i].x,point[i].y=-tmp[i].y,point[i].id=i; prepare();
sort(edge+,edge+m+,cmp2);
for (int i=;i<=n;i++) fa[i]=i;
for (int i=,cnt=n;i<=m&&cnt>k;i++) if (find(edge[i].u)!=find(edge[i].v))
cnt--,ans=max(ans,edge[i].c),fa[find(edge[i].u)]=find(edge[i].v);
printf("%d\n",ans);
return ;
}

 

老oj3444 && Pku3241 Object Clustering的更多相关文章

  1. POJ 3241 Object Clustering 曼哈顿最小生成树

    Object Clustering   Description We have N (N ≤ 10000) objects, and wish to classify them into severa ...

  2. 【Poj3241】Object Clustering

    Position: http://poj.org/problem?id=3241 List Poj3241 Object Clustering List Description Knowledge S ...

  3. poj 3241 Object Clustering (曼哈顿最小生成树)

    Object Clustering Time Limit: 2000MS   Memory Limit: 131072K Total Submissions: 2640   Accepted: 806 ...

  4. POJ 3241 Object Clustering(Manhattan MST)

    题目链接:http://poj.org/problem?id=3241 Description We have N (N ≤ 10000) objects, and wish to classify ...

  5. 【POJ 3241】Object Clustering 曼哈顿距离最小生成树

    http://poj.org/problem?id=3241 曼哈顿距离最小生成树模板题. 核心思想是把坐标系转3次,以及以横坐标为第一关键字,纵坐标为第二关键字排序后,从后往前扫.扫完一个点就把它插 ...

  6. 【poj3241】 Object Clustering

    http://poj.org/problem?id=3241 (题目链接) MD被坑了,看到博客里面说莫队要写曼哈顿最小生成树,我就写了一个下午..结果根本没什么关系.不过还是把博客写了吧. 转自:h ...

  7. POJ3241 Object Clustering 曼哈顿最小生成树

    题意:转换一下就是求曼哈顿最小生成树的第n-k条边 参考:莫涛大神的论文<平面点曼哈顿最小生成树> /* Problem: 3241 User: 96655 Memory: 920K Ti ...

  8. POJ3241 Object Clustering(最小生成树)题解

    题意:求最小生成树第K大的边权值 思路: 如果暴力加边再用Kruskal,边太多会超时.这里用一个算法来减少有效边的加入. 边权值为点间曼哈顿距离,那么每个点的有效加边选择应该是和他最近的4个象限方向 ...

  9. POJ 3241 曼哈顿距离最小生成树 Object Clustering

    先上几个资料: 百度文库有详细的分析和证明 cxlove的博客 TopCoder Algorithm Tutorials #include <cstdio> #include <cs ...

随机推荐

  1. 转 jquery 学习笔记

    jQ通过选择器选择元素,选择器的语法和css类似$(css选择器语法) 参数可以是id.class.tag等等通过如上选择就可以获得一个元素 jQuery名字冲突 解决方法: var jq=jQuer ...

  2. ecshop格式化商品价格

    <?php /** * 格式化商品价格 * * @access public * @param float $price 商品价格 * @return string */ function pr ...

  3. c# 发送邮件、附件 分类: C# 2014-12-17 16:41 201人阅读 评论(0) 收藏

    WinForm窗体代码如下: <span style="font-size:14px;">using System; using System.Collections. ...

  4. docker-compose.yml 语法说明

    YAML 模板文件语法 默认的模板文件是 docker-compose.yml,其中定义的每个服务都必须通过 image 指令指定镜像或 build 指令(需要 Dockerfile)来自动构建. 其 ...

  5. iOS--为视图添加阴影

    iOS–为视图添加阴影 情况一:视图添加圆角,在添加阴影 //阴影视图 self.viewShadow = [[UIView alloc]initWithFrame:CGRectMake(0, 0, ...

  6. warning:This application is modifying the autolayout engine from a background thread

    警告提示:This application is modifying the autolayout engine from a background thread, which can lead to ...

  7. 【python之路8】python基本数据类型(二)

    基本数据类型 4.列表(list) 创建列表 name_list = ['zhao','qian','sun','li'] 基本操作 索引 print(name_list[0]) #返回zhao pr ...

  8. 浅谈inline-block

    一.区分block,inline,inline-block 1.block block元素会独占一行,多个block元素会各自新起一行.默认情况下,block元素宽度自动填满其父元素宽度. block ...

  9. 点击其它地方隐藏div/事件冒泡/sweet-alert阻止冒泡

    点击document时把div隐藏,但点击div时阻止点击事件冒泡到document,从而实现“点击文档其它地方隐藏div,点击div本身不隐藏”.js代码如下:$("#div") ...

  10. 试着开发chrome插件

    我的第一个chrome插件,是app形式的 代码如下 创建一个文件: 1.manifest.json { "version": "1.0", "man ...