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