bzoj1604 / P2906 [USACO08OPEN]牛的街区Cow Neighborhoods

时间:2023-03-08 23:41:46
bzoj1604 / P2906 [USACO08OPEN]牛的街区Cow Neighborhoods

P2906 [USACO08OPEN]牛的街区Cow Neighborhoods

考虑维护曼哈顿距离:$\left | x_{1}-x_{2} \right |+\left | y_{1}-y_{2} \right |$

看起来很难维护的样子,我们尝试转化

设两个点$(x_{1},y_{1}),(x_{2},y_{2})  (x_{1}>=x_{2})$

那么它们的曼哈顿距离有2种情况

1.$y_{1}>y_{2}:\left | x_{1}-x_{2} \right |+\left | y_{1}-y_{2} \right |=x_{1}-x_{2}+y_{1}-y_{2}=(x_{1}+y_{1})-(x_{2}+y_{2})$

2.$y_{1}<y_{2}:\left | x_{1}-x_{2} \right |+\left | y_{1}-y_{2} \right |=x_{1}-x_{2}-y_{1}+y_{2}=(x_{1}-y_{1})-(x_{2}-y_{2})$

于是我们就可以转为维护$(X=x+y,Y=x-y)$

这样曼哈顿距离就愉快地转化为$max(X_{1}-X_{2},Y_{1}-Y_{2})$了

我们先把所有坐标按$x$升序排一遍。

蓝后$x$坐标用一个队列维护

$y$坐标则用$multiset$维护(当然你愿意的话也可以打个平衡树(逃))

每次在$multiset$里搞搞啥$lower\_bound$操作就行了。

至于点的联通问题,搞一个并查集

end.

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<set>
#define re register
using namespace std;
typedef long long ll;
void read(ll &x){
char c=getchar();x=;
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=(x<<)+(x<<)+(c^),c=getchar();
}
int max(int &a,int &b){return a>b?a:b;}
#define N 100002
struct data{
ll x,y;int id;
data(){}
data(ll A,ll B,int C):
x(A),y(B),id(C)
{}
bool operator < (const data &tmp) const{return y<tmp.y;}
}a[N];
bool cmp(const data &A,const data &B){return A.x<B.x;}
multiset <data> s;
int n,t,mxd=-1e9,fa[N],siz[N]; ll c;
int found(int x){return fa[x]==x?x:fa[x]=found(fa[x]);}
void uni(int x,int y){
int r1=found(x),r2=found(y);
if(r1!=r2){
siz[r1]+=siz[r2]; --t;
siz[r2]=; fa[r2]=r1;
}
}
int main(){
scanf("%d",&n); read(c); t=n; ll q1,q2;
for(re int i=;i<=n;++i){
read(q1); read(q2);
fa[i]=i; siz[i]=;
a[i]=data(q1+q2,q1-q2,i);
}sort(a+,a+n+,cmp);
s.insert(data(,1e16,));
s.insert(data(,-1e16,));//添加边界防止指针越界
s.insert(a[]); int hd=;
multiset<data>::iterator it;
for(re int i=;i<=n;++i){
while(a[hd].x+c<a[i].x) s.erase(s.find(a[hd++]));//队列维护
it=s.lower_bound(a[i]);//找到第一个>=a[i].y的
if((*it).y-a[i].y<=c) uni((*it).id,a[i].id);
if(a[i].y-(*(--it)).y<=c) uni((*it).id,a[i].id);
s.insert(a[i]);
}printf("%d ",t);
for(re int i=;i<=n;++i) mxd=max(mxd,siz[i]);
printf("%d",mxd);
return ;
}