Meeting point-2

Meeting point-2

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4312

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. 

It's an opportunity to show yourself in front of your predecessors!

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, all 8 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), (x-1,y+1), (x-1,y-1), (x+1,y+1), (x+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 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

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

20151438

Hint

In the first case, the meeting point is (0,2); the second is (0,0), the third is (1,-1) and the last is (-1,-1) 

 

在做完前一题的基础之上，后一题就稍微容易了一点。这题和上一题基本思路一样，就把哈夫曼距离改为切夫比雪距离就OK。先在有8个放向，如果把之前的那些点都旋转45度就原则上就变为一样的哈夫曼距离了。

简单的说就是Chebyshev距离，利用(x, y) -> (x - y, x+y) 转换为Manhattan距离，就和4311一样的代码了。

代码：

#include <stdio.h>#include <string.h>#include <math.h>#include <iostream>#include<algorithm>#define M 100005using namespace std;#define LL __int64struct point {    LL x;    LL y;    int id;} p[M];bool cmpx(point e1,point e2) {    return e1.x < e2.x;}bool cmpy(point e1,point e2) {    return e1.y < e2.y;}LL temp[M],sumx[M],sumy[M];int main() {    int cse;    cin>>cse;    while(cse--) {        int n;        scanf("%d",&n);        for(int i=1; i<=n; i++) {            scanf("%I64d%I64d",&p[i].x,&p[i].y);            LL tt=p[i].x;            p[i].x+=p[i].y;            p[i].y-=tt;            p[i].id = i;        }        memset(sumx,0,sizeof(sumx));        memset(sumy,0,sizeof(sumy));        sort(p+1, p+n+1, cmpx);        temp[0] = 0;        for(int i=1; i<=n; i++)            temp[i] = temp[i-1] + p[i].x;        for(int i=1; i<=n; i++)            sumx[ p[i].id ] = temp[n]  - 2*temp[i-1] + (2*i-n-2)*p[i].x;        sort(p+1, p+n+1, cmpy);        temp[0] = 0;        for(int i=1; i<=n; i++)            temp[i] = temp[i-1] + p[i].y;        for(int i=1; i<=n; i++)            sumy[ p[i].id ] = temp[n] -  2*temp[i-1] + (2*i-n-2)*p[i].y;        LL ans = sumx[1] + sumy[1];        for(int i=2; i<=n; i++) {            if(ans > sumx[i]+sumy[i])                ans = (sumx[i]+sumy[i]);        }        printf("%I64d\n",ans>>1);    }    return 0;}