HDOJ 4311 —— 排序&数学公式

Meeting point-1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2605    Accepted Submission(s): 814


Problem Description

It has been ten years since TJU-ACM established. And in this year all the 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 may be hard to find out where
to celebrate this meeting in order to minimize the sum travel time of all the retired TJU-ACMers. 

There is an infinite integer grid at which N retired TJU-ACMers have their houses on. They decide to unite at a common meeting place, which is someone's house. From any given cell, only 4 adjacent cells are reachable in 1 unit of time.

Eg: (x,y) can be reached from (x-1,y), (x+1,y), (x, y-1), (x, 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 an integer T represents there are T test cases. (0<T <=10)

For each test case, the first line is an integer n represents there are n retired TJU-ACMers. (0<n<=100000), the following n lines each contains two 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

4
6
-4 -1
-1 -2
2 -4
0 2
0 3
5 -2
6
0 0
2 0
-5 -2
2 -2
-1 2
4 0
5
-5 1
-1 3
3 1
3 -1
1 -1
10
-1 -1
-3 2
-4 4
5 2
5 -4
3 -1
4 3
-1 -2
3 4
-2 2

 

Sample Output

26
20
20
56
Hint
In the first case, the meeting point is (-1,-2); the second is (0,0), the third is (3,1) and the last is (-2,2)

 

Author

TJU

 

Source

2012 Multi-University Training Contest 2

 

Recommend

zhuyuanchen520

 
分别按x，y排序，求出dx和dy，然后取dx+dy的最小值即可。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>

using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
//#define mid ((l + r) >> 1)
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 100000 + 50;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x3f3f3f3f;
const int IMIN = 0x80000000;
const double E = 2.718281828;
#define eps 1e-8
#define DEBUG 1
#define mod 100000007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
//    #pragma comment(linker, "/STACK:102400000,102400000")__int64 a[10050];
int n;
struct Node
{
LL x , y , dx , dy;
}a[MAXN];
bool cmpx(Node a , Node b)
{
if(a.x == b.x)return a.y < b.y;
return a.x < b.x;
}
bool cmpy(Node a , Node b)
{
if(a.y == b.y)return a.x < b.x;
return a.y < b.y;
}

int main()
{
int t;
cin >> t;
while(t--)
{
scanf("%d" , &n);
FOR(i , 0 , n)
{
scanf("%I64d%I64d" , &a[i].x , &a[i].y);
a[i].dx = a[i].dy = 0;
}
sort(a,a  + n, cmpx);
FOR(i , 1 , n)a[0].dx += a[i].x - a[0].x;
FOR(i, 0 , n -1)a[i + 1].dx = a[i].dx - (n - 2 * i - 2) * (a[i + 1].x - a[i].x);
sort(a, a + n , cmpy);
FOR(i , 1 , n)a[0].dy += a[i].y - a[0].y;
FOR(i , 0 , n - 1)a[i + 1].dy = a[i].dy - (n - 2 * i - 2) * (a[i + 1].y - a[i].y);
FOR(i ,0 ,n)a[i].dx += a[i].dy;
LL ans  = a[0].dx;
FOR(i ,1 , n)
{

if(ans > a[i].dx)ans = a[i].dx;
}
printf("%I64d\n" , ans);
}
return 0;
}