BZOJ 3170 & 切比雪夫距离

题意:

　　给出N个点,在这N个点中选一个点使其它的点与这个点的切比雪夫距离和最小.

SOL:

　　TJOI真是...厚道还是防水...这种题目如果知道切比雪夫距离是什么那不就是傻逼题...如果不知道那不就懵逼了么...

　　与随意选点不同,这种给定点中选与x,y轴还是有一定关系的.所以我们要像树上搞那什么最小距离一样搞这种东西.然后就....

CODE:

　　

/*==========================================================================
# Last modified: 2016-03-04 19:52
# Filename:		3170.cpp
# Description: 
==========================================================================*/
#define me AcrossTheSky 
#include <cstdio> 
#include <cmath> 
#include <ctime> 
#include <string> 
#include <cstring> 
#include <cstdlib> 
#include <iostream> 
#include <algorithm> 
  
#include <set> 
#include <map> 
#include <stack> 
#include <queue> 
#include <vector> 
 
#define lowbit(x) (x)&(-x) 
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++) 
#define FORP(i,a,b) for(int i=(a);i<=(b);i++) 
#define FORM(i,a,b) for(int i=(a);i>=(b);i--) 
#define ls(a,b) (((a)+(b)) << 1) 
#define rs(a,b) (((a)+(b)) >> 1) 
#define getlc(a) ch[(a)][0] 
#define getrc(a) ch[(a)][1] 
 
#define maxn 130000 
#define maxm 130000 
#define pi 3.1415926535898 
#define _e 2.718281828459 
#define INF 1070000000 
using namespace std; 
typedef long long ll; 
typedef unsigned long long ull; 
 
template<class T> inline 
void read(T& num) { 
    bool start=false,neg=false; 
    char c; 
    num=0; 
    while((c=getchar())!=EOF) { 
        if(c=='-') start=neg=true; 
        else if(c>='0' && c<='9') { 
            start=true; 
            num=num*10+c-'0'; 
        } else if(start) break; 
    } 
    if(neg) num=-num; 
} 
/*==================split line==================*/ 
struct Infor{
	double x,y;
	int id;
}p[maxn];
double ans[maxn];
int cmpx(Infor a,Infor b){return a.x<b.x;}
int cmpy(Infor a,Infor b){return a.y<b.y;}
int main(){ 
	int n; read(n); 
	double sumx=0,sumy=0;
	FORP(i,1,n) {
		double x,y;
		scanf("%lf%lf",&x,&y);
		p[i].x=(x+y)/2.0;	p[i].y=(x-y)/2.0;
		p[i].id=i;
		sumx+=p[i].x; sumy+=p[i].y;
	}
	sort(p+1,p+1+n,cmpx);
	double temp=0;
	FORP(i,1,n){
		//int t=p[i].x-p[i-1].x;
		ans[p[i].id]+=((i-1)*p[i].x)-temp;
		ans[p[i].id]-=((n-i+1)*p[i].x-(sumx-temp));
		temp+=p[i].x;
	}
	sort(p+1,p+1+n,cmpy);

	temp=0;
	FORP(i,1,n){
		ans[p[i].id]+=(i-1)*p[i].y-temp;
		ans[p[i].id]-=((n-i+1)*p[i].y-(sumy-temp));
		temp+=p[i].y;
	}
	double out=ans[1];
	FORP(i,1,n) out=min(out,ans[i]);
	printf("%.0lf
",out);
}