HDU 4311 Meeting point-1 简单几何

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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

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 is (3,1) and the last is (-2,2)
题意：给你一些点，找到一点使其他点到该点的曼哈顿距离最小，（曼哈顿距离是指只有垂直和竖直方向）
思路：直接暴力求解，遍历各个点，依次把他们作为所求点，再找最小值即可，data[i].x,data[i].y为第i个点的坐标。data[i].sum存储其他点到该点的曼哈顿距离，找最小值即可，那只要确定个点到该点的曼哈顿距离就行了
data[i].sum=(i-1)*data[i].x-(data[1].x~data[i-1].x累加)+（data[i+1].x~data[n].x累加）-(n-i)*data[i].x
这是水平方向的距离
竖直方向同理
然后找到最小值即可：
ac代码
#include <iostream>#include <cstdio>#include <algorithm>using namespace std;struct point{    long long x;    long long y;    long long sum;}data[100001];bool cmpx(point a,point b){    return a.x<b.x;}bool cmpy(point a,point b){    return a.y<b.y;}int main(){    int t,d;    scanf("%d",&t);    while(t--)    {        scanf("%d",&d);        for(int i=1;i<=d;i++)        scanf("%lld%lld",&data[i].x,&data[i].y);        sort(data+1,data+d+1,cmpx);        long long sum1=0;        for(int i=1;i<=d;i++)        {            data[i].sum=(i-1)*data[i].x-sum1;            sum1+=data[i].x;        }        sum1=0;        for(int i=d;i>=1;i--)        {            data[i].sum+=sum1-(d-i)*data[i].x;            sum1+=data[i].x;        }        sort(data+1,data+d+1,cmpy);        sum1=0;        for(int i=1;i<=d;i++)        {            data[i].sum+=(i-1)*data[i].y-sum1;            sum1+=data[i].y;        }        sum1=0;        for(int i=d;i>=1;i--)        {            data[i].sum+=sum1-(d-i)*data[i].y;            sum1+=data[i].y;        }        long long ans=data[1].sum;        for(int i=2;i<=d;i++)        ans=min(ans,data[i].sum);        printf("%lld\n",ans);    }    return 0;}
当时做这道题时一直不过，后来才发现犯了个zz的错误，在结构体外面定义了一个sum[100]储存结果但是忽略了一旦排序后，就不再对应原来那个数的储存值了

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