HDU 4311 Prefixes and

Description

It has been ten years since TJU-ACM established. Retired tju-acmers want to get together to celebrate the tenth anniversary. Because the retired tju-acmers may live in different places around the world, it could be hard-to-find out where to Celebrat E This meeting in order to minimize the sum travel time of the the retired tju-acmers.
There is a infinite integer grid at which N retired Tju-acmers has their houses on. They decide to unite at a common meeting place, which was someone ' s house. From any given cell, only 4 adjacent cells is reachable in 1 unit of time.
Eg: (x, Y) can is reached from (X-1,y), (X+1,y), (x, Y-1), (×, y+1).
Finding a common meeting place which minimizes the sum of the travel time of all the retired tju-acmers.

Input

The first line is a integer t represents there is t test cases. (0<t <=10)
For each test case, the first line is a integer n represents there is n retired tju-acmers. (0<n<=100000), the following n lines each contains a integers x, y coordinate of the i-th Tju-acmer. ( -10^9 <= x, y <= 10^9)

Output

For each test case, output the minimal sum of travel times.

Sample Input

46-4-1-1-22-40 20 35-260 02 0-5 -22-2-1 24 05-5 1-1 33 13-11-110-1-1-3 2-4 45 25-43-14 3-1-23 4-2 2

Sample Output

26202056         

Hint

In the first case, the meeting point is ( -1,-2); The second is (0,0), the third are (3,1) and the Last is ( -2,2) test instructions: N people in different places, choose one of the rooms, so that other people go to the distance of this place and the shortest problem: the distance from the topic is Manhattan distance | x1-x2|+|y1-y2|; So you can consider the x, y, separate calculation, and finally take the max; How to quickly handle the distance from other positions to I position dis[i]? First, the coordinates are ordered to record the prefix and the distance between the two i,j=i+1 that can be found in the adjacent position dis[j]=dis[i]+ (j-2) * (Sumx[j]-sumx[i])-(N-J) * (Sumx[j]-sumx[i]) =dis[i]+ ( I-N) * (Sumx[j]-sumx[i])

1 /******************************2 code by drizzle3 blog:www.cnblogs.com/hsd-/4 ^ ^    ^ ^5 o o6 ******************************/7 //#include <bits/stdc++.h>8#include <iostream>9#include <cstring>Ten#include <cstdio> One#include <map> A#include <algorithm> -#include <cmath> - #definell Long Long the #definePI ACOs (-1.0) - #defineMoD 1000000007 - intT; - using namespacestd; + structnode - { + ll X,y,pos; A} n[100005]; atll sumx[100005]; -ll sumy[100005]; -ll ans1[100005]; -ll ans2[100005]; - intEXM; - BOOLCMP1 (structNode AA,structnode BB) in { -     returnaa.x<bb.x; to } + BOOLCMP2 (structNode AA,structnode BB) - { the     returnaa.y<bb.y; * } $ intMain ()Panax Notoginseng { -      while(~SCANF ("%d",&t)) the     { +          for(intI=1; i<=t; i++) A         { thescanf"%d",&EXM); +memset (N,0,sizeof(N)); -memset (ANS1,0,sizeof(ans1)); $memset (Ans2,0,sizeof(ANS2)); $              for(intj=1; j<=exm; J + +) -             { -scanf"%i64d%i64d",&n[j].x,&n[j].y); then[j].pos=J; -             }WuyiSort (n+1, n+1+exm,cmp1); thesumx[1]=0; -              for(intj=2; j<=exm; J + +) Wu             { -sumx[j]=sumx[j-1]+n[j].x-n[j-1].x; Aboutans1[n[1].pos]+=Sumx[j]; $             } -              for(intj=2; j<=exm; J + +) -             { -  Aans1[n[j].pos]=ans1[n[j-1].pos]-(EXM-J) * (sumx[j]-sumx[j-1]) + (J-2) * (sumx[j]-sumx[j-1]); +             } theSort (n+1, n+1+exm,cmp2); -sumy[1]=0; $              for(intj=2; j<=exm; J + +) the             { thesumy[j]=sumy[j-1]+n[j].y-n[j-1].y; theans2[n[1].pos]+=Sumy[j]; the             } -              for(intj=2; j<=exm; J + +) in             { theans2[n[j].pos]=ans2[n[j-1].pos]-(EXM-J) * (sumy[j]-sumy[j-1]) + (J-2) * (sumy[j]-sumy[j-1]); the             } Aboutll maxn=ans1[1]+ans2[1]; the              for(intj=1; j<=exm; J + +) the             { the                 if(ans1[j]+ans2[j]<MAXN) +                 { -maxn=ans1[j]+Ans2[j]; the                 }Bayi             } theprintf"%i64d\n", MAXN); the         } -     } -     return 0; the}

HDU 4311 Prefixes and