HDU-4311 meeting point-1 Manhattan distance quick computing

This question has a certain skill, need to calculate the meaning of a point in the time complexity of nlog (N), N-1 points to the distance between the sum, here need to use such a technique, calculate the distance from X and Y separately. First, calculate the distance from the X axis, then sort the order first, and then sum the distance to the point P (P-1) * XP-sum (x1... xp-1) + sum (XP + 1... XN)-(N-P) * XP; similarly, the distance from the Y axis can be calculated, which is accumulated to a struct. So you can directly find the minimum value.

The Code is as follows:

#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;typedef long long int Int64;int N;struct Point{    Int64 x, y, sum;}e[100005];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;    Int64 sum, Min;    scanf("%d", &T);    while (T--) {        sum = 0;        Min = 1LL << 62;        scanf("%d", &N);        for (int i = 1; i <= N; ++i) {            e[i].sum = 0;            scanf("%I64d %I64d", &e[i].x, &e[i].y);        }        sort(e+1, e+1+N, cmpx);        for (int i = 1; i <= N; ++i) {            e[i].sum += (i-1) * e[i].x - sum;            sum += e[i].x;        }        sum = 0;        for (int i = N; i >= 1; --i) {            e[i].sum += sum - (N-i) * e[i].x;            sum += e[i].x;        }        sum = 0;        sort(e+1, e+1+N, cmpy);        for (int i = 1; i <= N; ++i) {            e[i].sum += (i-1) * e[i].y -sum;            sum += e[i].y;        }        sum = 0;        for (int i = N; i >= 1; --i) {            e[i].sum += sum - (N-i) * e[i].y;            Min = min(Min, e[i].sum);            sum += e[i].y;        }        printf("%I64d\n", Min);    }    return 0;}