csu1896——Symmetry

We call a figure made of points is left-right symmetric as it is possible to fold the sheet of paper along a vertical line and to cut the figure into two identical halves.For example, if a figure exists five points, which respectively are (-2,5),(0,0),(2,3),(4,0),(6,5). Then we can find a vertical line x = 2 to satisfy this condition. But in another figure which are (0,0),(2,0),(2,2),(4,2), we can not find a vertical line to make this figure left-right symmetric.Write a program that determines whether a figure, drawn with dots, is left-right symmetric or not. The dots are all distinct.
Input

The input consists of T test cases. The number of test cases T is given in the first line of the input file. The first line of each test case contains an integer N, where N (1 <=N <= 1, 000) is the number of dots in a figure. Each of the following N lines contains the x-coordinate and y-coordinate of a dot. Both x-coordinates and y-coordinates are integers between −10, 000 and 10, 000, both inclusive.
Output

Print exactly one line for each test case. The line should contain 'YES' if the figure is left-right symmetric,and 'NO', otherwise.
Sample Input

3
5
-2 5
0 0
6 5
4 0
2 3
4
2 3
0 4
4 0
0 0
4
5 14
6 10
5 10
6 14

Sample Output

YES
NO

YES

题意：给你一些点，让你去找一条垂直于x轴的直线，使得所有点关于他对称。

解题思路：将这些点排序，先按照纵坐标的从小到大，在安装横坐标从小到大。在第一批纵坐标相等的点中，找到这条直线。然后再用这条直线去验证其他的点。

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;const int N=1005;const double eps=1e-8;//必须小于等于他struct Node{    int x,y;} p[N];bool cmp(Node a,Node b){    if(a.y!=b.y)        return a.y<b.y;    return a.x<b.x;}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0; i<n; i++)            scanf("%d%d",&p[i].x,&p[i].y);        sort(p,p+n,cmp);        int l,r;        double m;        l=p[0].x,r=p[0].x;        int j=1;        while(j<n)        {            if(p[0].y!=p[j].y)            {                break;            }            l=min(l,p[j].x);            r=max(r,p[j].x);            j++;        }        m=(l+r)/2.0;//找到对称轴       // cout<<m<<endl;        int flag=0;        for(int i=0;i<n;i++)        {            int j=i+1;            while(p[j].y==p[i].y)//找到纵坐标相等的所有点            {                j++;            }            int l=i,r=j-1;           // cout<<l<<" "<<r<<endl;            while(l<=r)            {                double temp=(p[l].x+p[r].x)/2.0;                if(fabs(temp-m)>eps)//判断前后两个点的对称轴是否为m                {                    flag=1;                    //cout<<"flag"<<l<<" "<<r<<endl;                    break;                }                l++;                r--;            }            if(flag)                break;            i=j-1;        }        if(flag)            printf("NO\n");        else            printf("YES\n");    }    return 0;}