UVA 10585 - Center of Symmetry

Description

Given is a set of n points with integer coordinates. Your task is to decide whether the set has a center of symmetry.

A set of points S has the center of symmetry if there exists a point s (not necessarily in S) such that for every point p in S there exists a point q in S such that p-s = s-q.

Input

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines contain two integer numbers each which are the x and y coordinates of a point. Every point is unique and we have that -10000000 <= x, y <= 10000000.

Output

For each set of input data print yes if the set of points has a center of symmetry and no otherwise.

Sample Input

181 103 66 86 23 -41 0-2 -2-2 4

Sample Output

yes

题意：给出一个 T 表示有 T 组输入数据，每个测试数据有一个 N，表示有 N 个点，每个点给出x和y，问这个图形是否关于一点对称。

通过排序，然后计算出每对点的中点，只要所有的点的中点都是一样就可以是yes。

#include <cstdio>#include <algorithm>using namespace std;struct node{    int x, y;};node point[10005];int cmp(node a, node b){    if (a.x == b.x)        return a.y < b.y;    return a.x < b.x;}int main(){    int T;    int n;    scanf("%d", &T);    while (T--)    {        scanf("%d", &n);        for (int i = 0; i < n; ++i)            scanf("%d%d", &point[i].x, &point[i].y);        sort(point, point + n, cmp);        int flag = 1;        int sym_x = point[0].x + point[n-1].x;        int sym_y = point[0].y + point[n-1].y;        for (int i = 1; i < n/2; ++i)        {            int j = n - i -1;            if (point[i].x + point[j].x != sym_x || point[i].y + point[j].y != sym_y)            {                flag = 0;                break;            }        }        if (flag == 1) printf("yes\n");        else printf("no\n");    }    return 0;}