BZOJ 1604: [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居

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题目

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1604: [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居

Time Limit: 5 Sec  Memory Limit: 64 MB

Description

了解奶牛们的人都知道，奶牛喜欢成群结队．观察约翰的N(1≤N≤100000)只奶牛，你会发现她们已经结成了几个“群”．每只奶牛在吃草的时候有一个独一无二的位置坐标Xi，Yi(l≤Xi，Yi≤[1．.10^9]；Xi，Yi∈整数．当满足下列两个条件之一，两只奶牛i和j是属于同一个群的：

  1．两只奶牛的曼哈顿距离不超过C(1≤C≤10^9)，即lXi - xil+IYi - Yil≤C.

  2．两只奶牛有共同的邻居．即，存在一只奶牛k，使i与k，j与k均同属一个群．

    给出奶牛们的位置，请计算草原上有多少个牛群，以及最大的牛群里有多少奶牛

Input

   第1行输入N和C，之后N行每行输入一只奶牛的坐标．

Output

仅一行，先输出牛群数，再输出最大牛群里的牛数，用空格隔开．

Sample Input

4 2
1 1
3 3
2 2
10 10

* Line 1: A single line with a two space-separated integers: the
number of cow neighborhoods and the size of the largest cow
neighborhood.

Sample Output

2 3

OUTPUT DETAILS:
There are 2 neighborhoods, one formed by the first three cows and
the other being the last cow. The largest neighborhood therefore
has size 3.

 

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题解

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这题很明显就是再考并查集！【真的吗，你看下数据范围！】关于曼哈顿距离的技巧，将每个点变成(x+y,x-y)这样两个点之间的曼哈顿距离就是|x1-x2|+|y1-y2|，这样就可以维护一个平衡树，根据前驱和后继上面的节点来维护并查集求得答案了！

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代码

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/*Author:WNJXYK*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
using namespace std;

#define LL long long
#define Inf 2147483647
#define InfL 10000000000LL

inline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;}
inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;}
inline int remin(int a,int b){if (a<b) return a;return b;}
inline int remax(int a,int b){if (a>b) return a;return b;}
inline LL remin(LL a,LL b){if (a<b) return a;return b;}
inline LL remax(LL a,LL b){if (a>b) return a;return b;}

inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,c,ans,mx;
int fa[100005],tot[100005];
struct data{LL x,y;int id;}a[100005];
multiset <data> b;
set <data>::iterator it;
inline bool operator<(data a,data b){
    return a.y<b.y;
}
inline bool cmpx(data a,data b){
    return a.x<b.x;
}
int find(int x){
    return x==fa[x]?x:fa[x]=find(fa[x]);
}
inline void un(int x,int y){
    int p=find(x),q=find(y);
    if(p!=q){
        fa[p]=q;
        ans--;
    }
}
void solve(){
    b.insert((data){0,InfL,0});b.insert((data){0,-InfL,0});   
    int now=1;b.insert(a[1]);
    for(int i=2;i<=n;i++){
        while(a[i].x-a[now].x>c){
            b.erase(b.find(a[now]));
            now++;
        }
        it=b.lower_bound(a[i]);
        data r=*it,l=*--it;
        if(a[i].y-l.y<=c)
            un(a[i].id,l.id);
        if(r.y-a[i].y<=c)
            un(a[i].id,r.id);
        b.insert(a[i]);
    }
}
int main(){
    n=read();c=read();ans=n;
    for(int i=1;i<=n;i++)fa[i]=i;
    for(int i=1;i<=n;i++){
        int t1=read(),t2=read();
        a[i].x=t1+t2,a[i].y=t1-t2;a[i].id=i;
    }
    sort(a+1,a+n+1,cmpx);
    solve();
    for(int i=1;i<=n;i++)
        tot[find(i)]++;
    for(int i=1;i<=n;i++)
        mx=max(mx,tot[i]);
    printf("%d %d\n",ans,mx);
    return 0;
}