[hdu4311]Meeting point-1

题意：在整数坐标轴上找一个距离所有给定点距离最小的点。

解题关键：对x和y分别处理，前缀和预处理所有点到最小点的距离，每点的$sum$等于左边的贡献+右边的贡献，最后取$min$即可。

复杂度：$O(nlogn)$

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<cmath>
using namespace std;
typedef long long ll;
const ll INF = 1ll<<61-1;
struct node{
    ll x,y,sum;
}p[100002];
bool cmp1(const node &a,const node &b){
    return a.x<b.x;
}

bool cmp2(const node &a,const node &b){
    return a.y<b.y;
}
ll dis[100002];
ll sum1[100002];
int main(){
    ll t;
    scanf("%I64d",&t);
    while(t--){
        memset(p,0,sizeof p);
        ll n;
        scanf("%I64d",&n);
        for(int i=1;i<=n;i++)   scanf("%I64d%I64d",&p[i].x,&p[i].y);
        sort(p+1,p+n+1,cmp1);
        for(int i=1;i<=n;i++) dis[i]=p[i].x-p[1].x;
        for(int i = 1;i <= n;i++) sum1[i]=sum1[i-1]+dis[i];
        for(int i=1;i<=n;i++){
            p[i].sum+=abs((i-1)*dis[i]-sum1[i-1]);
            p[i].sum+=abs((sum1[n]-sum1[i])-dis[i]*(n-i));
        }
        sort(p+1,p+n+1,cmp2);
        for(int i=1;i<=n;i++) dis[i]=p[i].y-p[1].y;
        for(int i = 1;i <= n;i++) sum1[i]=sum1[i-1]+dis[i];
        for(int i=1;i<=n;i++){
            p[i].sum+=abs((i-1)*dis[i]-sum1[i-1]);
            p[i].sum+=abs((sum1[n]-sum1[i])-dis[i]*(n-i));
        }
        ll ma=INF;
        for(int i=1;i<=n;i++){
            if(p[i].sum<ma){
                ma=p[i].sum;
            }
        }
        printf("%lld
",ma);
    }
    
}