负载平衡Load Balancing_Silver

题目描述

Farmer John's $(x_1, y_1) \ldots (x_n, y_n)(x1,y1)…(xn,yn) on his two-dimensional farm (1 \leq N \leq 10001≤N≤1000, and the x_ixi's and y_iyi's are positive odd integers of size at most 1,000,0001,000,000). FJ wants to partition his field by building a long (effectively infinite-length) north-south fence with equation x=ax=a (aa will be an even integer, thus ensuring that he does not build the fence through the position of any cow). He also wants to build a long (effectively infinite-length) east-west fence with equation y=by=b, where bb is an even integer. These two fences cross at the point (a,b)(a,b), and together they partition his field into four regions.$

FJ wants to choose

smallest possible value for

给你一个矩阵，里面有些点，让你横向切一刀，纵向切一刀，使得得到的四个区域内的最大的点数最少。

输入格式

The first line of the input contains a single integer,

each contain the location of a single cow, specifying its

coordinates.

输出格式

You should output the smallest possible value of

positioning his fences optimally.

输入输出样例

输入 #1

输出 #1

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;

void chmin(int &x,int y){
    if(x>y)x=y;
}

const int N=40000005;
int n,i,u;
int px[N],py[N];
int id[N];

bool hhh(int x,int y){
   return py[x]<py[y];
}

bool kkk(int x,int y){
   return px[x]<px[y];
}

int lisan_py(){
    for(i=1;i<=n;++i){
        id[i]=i;
    }
    sort(id+1,id+n+1,hhh);
    int now=0,top=0;
    for(i=1;i<=n;++i){
        int x=id[i];
        if(py[x]!=now){
            now=py[x];
            ++top;
        }
        py[x]=top;
    }
    return top;
}

#define cl (i<<1)
#define cr (cl+1)

int al[N*4],ar[N*4],d;

void init(int *a){
    for(i=1;i<=n;++i){
        ++a[d+py[i]];
    }
    for(i=u+d>>1;i;--i){
        a[i]=a[cl]+a[cr];
    }
}

void add(int *a,int i,int w){
    for(i+=d;i;i>>=1){
        a[i]+=w;
    }
}

int erfen(){
    i=1;
    int all=0,alr=0,arl=0,arr=0;
    while(i<=d){
        int mxl=max(all+al[cl],arl+ar[cl]),mxr=max(alr+al[cr],arr+ar[cr]);
        if(mxl<=mxr){
            all+=al[cl];
            arl+=ar[cl];
            i=cr;
        }
        else{
            alr+=al[cr];
            arr+=ar[cr];
            i=cl;
        }
    }
    int mxl=max(max(all+al[i],arl+ar[i]),max(all,arr)),
    mxr=max(max(alr+al[i],arr+al[i]),max(all,arl));
    return min(mxl,mxr); 
}


int main(){
    int i;
    scanf("%d",&n);
    for(i=1;i<=n;++i){
        scanf("%d%d",px+i,py+i);
    }
    u=lisan_py();
    for(d=1;d<u;d<<=1);d-=1;
    init(ar);
    sort(id+1,id+n+1,kkk);
    int ans=n;
    for(i=1;i<=n;++i){
        int x=id[i];
        add(ar,py[x],-1);
        add(al,py[x],1);
        chmin(ans,erfen());
    }
    printf("%d\n",ans);
}