【bzoj1604/Usaco2008 Open】Cow Neighborhoods 奶牛的邻居——平衡树+并查集

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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题解参考hzwer的->戳这里

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#define mem(a,p) memset(a,p,sizeof(a))
typedef long long LL;
const LL inf=1e18;
const int N=1e5+10;
using std::sort;
using std::max;
using std::abs;
struct node{LL x,y;int id;}e[N];
bool operator <(node a,node b){return a.y<b.y;}   
typedef std::multiset<node>me;
typedef me::iterator IT;
int n,c,mx=0,ton[N],fa[N];
me mset;
bool cmp(node aa,node bb){return aa.x<bb.x;}
int read(){
    int ans=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-48;c=getchar();}
    return ans*f;
}
int ans;
int find(int x){
    if(x==fa[x])return x;
    fa[x]=find(fa[x]);
    return fa[x];
}
void merge(int x,int y){
    x=find(x);y=find(y);
    if(x!=y){
        fa[y]=x;ans--;
    }
}
void work(){
    int h=1;mset.insert((node){0,inf,0});mset.insert((node){0,-inf,0});
    mset.insert(e[1]);
    for(int i=2;i<=n;i++){
        while(h<i&&e[i].x-e[h].x>c){
            IT it=mset.find(e[h]);
            mset.erase(it);
            h++;
        }
        IT it=mset.lower_bound(e[i]),it1=--it;it++;
        if(it1->y>=e[i].y-c)merge(e[i].id,it1->id);
        if(it->y<=e[i].y+c)merge(e[i].id,it->id);
        mset.insert(e[i]);
    }
}
int main(){
    n=read();c=read();ans=n;
    for(int i=1,x,y;i<=n;i++){
        x=read();y=read();
        e[i].x=x+y;e[i].y=x-y;e[i].id=i;
        fa[i]=i;
    }
    sort(e+1,e+1+n,cmp);
    work();
    for(int i=1;i<=n;i++)ton[find(i)]++;
    for(int i=1;i<=n;i++)mx=max(mx,ton[i]);
    printf("%d %d\n",ans,mx);
    return 0;
}