HDU-4311-meeting point-1-Question

Given n coordinates, calculate the minimum value of one of the coordinates to the sum of other coordinates.

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4311

Solution:

1. sort by X. Sumx []. Sumx [I] indicates the sum of the X coordinates of the first I including I. And record the location ID when sorting by X. (Coordinates and, not distance and)

2. sort by Y. Sumy [I];

3. for each I, (P [I]. y) * (I)-Sumy [I] indicates the sum of the vertical distance between the vertex and the I point below the I point, (Sumy [N]-Sumy [I]-(P [I]. y) * (N-I) indicates the sum of the vertical distance between the vertex and the I point above the I point.

(P [I]. x) * (j)-sumx [J] indicates the sum of the horizontal distance between the left vertex and the I vertex of the I vertex, (sumx [N]-sumx [J]-(P [I]. x) * (N-j) indicates the sum of the horizontal distance between the vertices and I points on the right of the I point. (J is the subscript of I when sorting by X)

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#define maxn 100010typedef long long L;using namespace std;struct Node{    L x,y;    int id;} p[maxn];L sumx[maxn],sumy[maxn];bool cmpx(Node a , Node b){    return a.x < b.x;}bool cmpy(Node a, Node b){    return a.y < b.y;}int T, n;int main(){    scanf("%d",&T);    while(T --){        scanf("%d",&n);        for(int i = 1; i <= n; ++ i){            scanf("%I64d%I64d", &p[i].x, &p[i].y);        }        sort(p + 1, p + 1 + n, cmpx);        sumx[1] = p[1].x;p[1].id = 1;        for(int i = 2; i <= n; ++ i){            sumx[i] = sumx[i - 1] + p[i].x;            p[i].id = i;        }        sort(p + 1, p + 1 + n, cmpy);        sumy[1] = p[1].y;        for(int i = 2; i <= n; ++ i){            sumy[i] = sumy[i - 1] + p[i].y;        }        L ans = 1; ans <<= 60;//        for(int i = 1; i <= n; i ++){            int j = p[i].id;            L yy = ( p[i].y ) * (i) - sumy[i];            yy += (sumy[n] - sumy[i] - (p[i].y) * (n - i) );            L xx = ( p[i].x ) * (j) - sumx[j];            xx += (sumx[n] - sumx[j] - (p[i].x) * (n - j) );            xx += yy;            ans = min(xx,ans);        }        printf("%I64d\n", ans);    }    return 0;}

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2. sort by Y and calculate Sumy [].

3. for any point I, (P [I]. y) * (I)-Sumy [I] is the vertical distance and (Sumy [N]-Sumy [I]-(P [I]. y) * (N-I) is the vertical distance and.

At the same time: (P [I]. x) * (ID)-sumx [ID] indicates the horizontal distance and (sumx [N]-sumx [J]-(P [I]. x) * (N-j) is the horizontal distance and of the point on the right of the I point.

The coordinates are written incorrectly and cannot be called.