BZOJ 1941 [Sdoi2010]Hide and Seek

题解:每个点向四个方向分别求最远点和最近点,用树状数组维护即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=500009;
const int oo=1000000000;

int n;
int ans=oo;

struct Point{
	int x,y;
	int mindist,maxdist;
}p[maxn];
bool cmpx(const Point &rhs1,const Point &rhs2){
	if(rhs1.x==rhs2.x)return rhs1.y<rhs2.y;
	else return rhs1.x<rhs2.x;
}

inline int lowbit(int x){
	return x&(-x);
}

int b[maxn],nn;
struct FenwickTree{
	int c[maxn];
	
	void Addp(int x,int val){
		while(x<=nn){
			c[x]=max(c[x],val);
			x+=lowbit(x);
		}
	}
	int Query(int x){
		int ret=-oo;
		while(x){
			ret=max(ret,c[x]);
			x-=lowbit(x);
		}
		return ret;
	}
	void Fenwickinit(){
		for(int i=0;i<maxn;++i)c[i]=-oo;
	}
}T[3];

void Sol(){
	T[1].Fenwickinit();
	T[2].Fenwickinit();
	for(int i=1;i<=n;++i)b[i]=p[i].y;
	sort(b+1,b+1+n);
	nn=unique(b+1,b+1+n)-b-1;
	for(int i=1;i<=n;++i){
		int pla=lower_bound(b+1,b+1+nn,p[i].y)-b;
		int tm1=T[1].Query(pla);
		int tm2=-T[2].Query(pla);
		p[i].mindist=min(p[i].mindist,p[i].x+p[i].y-tm1);
		p[i].maxdist=max(p[i].maxdist,p[i].x+p[i].y-tm2);
		T[1].Addp(pla,p[i].x+p[i].y);
		T[2].Addp(pla,-p[i].x-p[i].y);
	}
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d%d",&p[i].x,&p[i].y);
		p[i].mindist=oo;
		p[i].maxdist=-oo;
	}
	
	sort(p+1,p+1+n,cmpx);
	Sol();
	
	for(int i=1;i<=n;++i)p[i].y=-p[i].y;
	sort(p+1,p+1+n,cmpx);
	Sol();
	
	for(int i=1;i<=n;++i)p[i].y=-p[i].y;
	for(int i=1;i<=n;++i)p[i].x=-p[i].x;
	sort(p+1,p+1+n,cmpx);
	Sol();
	
	for(int i=1;i<=n;++i)p[i].y=-p[i].y;
	sort(p+1,p+1+n,cmpx);
	Sol();
	
	for(int i=1;i<=n;++i){
		ans=min(ans,p[i].maxdist-p[i].mindist);
	}
	printf("%d\n",ans);
	return 0;
}

  

posted @ 2018-03-04 11:44  ws_zzy  阅读(69)  评论(0编辑  收藏  举报