POJ 2526 Center of symmetry(计算几何--对称点)

Center of symmetry

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2190		Accepted: 972

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

1

8

1 10

3 6

6 8

6 2

3 -4

1 0

-2 -2

-2 4

Sample Output

yes

Source

Alberta Collegiate Programming Contest 2003.10.18

 解题报告：这道题就行求能否找到一个中心点能使其他的每个点都有其对称点，其中中心点可以是题目所给的点，也可以假设一个中心点，把点的坐标排序即可！我求中点坐标的时候没有除以2，因为计算几何中能尽量不用除法，注意精度问题；

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX = 10005;
int n, t;
bool flag;
struct Point
{
    double x;
    double y;
}p[MAX], center;
int cmp(Point a, Point b)//先按x从大到小排序，若x相等时则按y从大到小排序
{
    if (a.x == b.x) return a.y > b.y;
    return a.x > b.x ;
}
bool Judge()
{
    int i;
    if (n % 2)
    {
        center.x = 2 * p[n / 2].x;
        center.y = 2 * p[n / 2].y;
    }
    else
    {
        center.x = p[n / 2].x + p[n / 2 - 1].x;
        center.y = p[n / 2].y + p[n / 2 - 1].y;
    }
    for (i = 0; i < n / 2; ++i)
    {
        if (p[i].x + p[n - i - 1].x != center.x || p[i].y + p[n - i - 1].y != center.y)
        {
            return false;
        }
    }
    return true;
}
int main()
{
    int i, j;
    scanf("%d", &t);
    while (t --)
    {
        scanf("%d", &n);
        for (i = 0; i < n; ++i)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
        }
        sort(p, p + n, cmp);
        flag = Judge();
        if (flag)
        {
            printf("yes\n");
        }
        else
        {
            printf("no\n");
        }
    }
    return 0;
}