『算法-ACM 竞赛-POJ』 3241Object Clustering 曼哈顿距离最小生成树

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

#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("%d\n",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;
}