对称轴

The figure shown on the left is left-right symmetric as it is possible to fold the sheet of paper along avertical line, drawn as a dashed line, and to cut the figure into two identical halves. The figure on theright is not left-right symmetric as it is impossible to find such a vertical line.Write a program that determines whether a figure, drawn with dots, is left-right symmetric or not.The dots are all distinct.InputThe 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 dotsin a figure. Each of the following N lines contains the x-coordinate and y-coordinate of a dot. Bothx-coordinates and y-coordinates are integers between −10, 000 and 10, 000, both inclusive.OutputPrint 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：

有T个测试样例，每个测试样例有n个点，判断由着n 个点构成的图形是否左右对称，对称输出YES,否则输出NO

#include<cstring>#include<cstdio>#include<algorithm>using namespace std;struct A{    int x,y;} f[1001],g[1001];bool cmp(A a,A b){    return a.x<b.x;}bool cmp1(A a,A b){    if(a.x!=b.x)        return a.x<b.x;    return a.y>b.y;}bool cmp2(A a,A b){    if(a.x!=b.x)        return a.x<b.x;    return a.y<b.y;}int main(){    int n,m;    scanf("%d",&n);    while(n--)    {        int num=0,ok=0;        int xx,yy,s=0,su=0;        scanf("%d",&m);        for(int i=0; i<m; i++)        {            scanf("%d %d",&g[i].x,&g[i].y);            if(i==0)            {                xx=g[i].x;                yy=g[i].y;            }            else            {                if(xx==g[i].x)                    s++;                if(yy==g[i].y)                    su++;            }        }        if(s==m-1||su==m-1)//当平行于x轴或者y轴的时候        {            printf("YES\n");            continue;        }        sort(g,g+m,cmp);//先按x坐标从小到大排序，小的一部分在左边，大的在右边        for(int i=0; i<m/2; i++)//找左边部分        {            f[i].x=g[i].x;            f[i].y=g[i].y;        }        sort(f,f+m/2,cmp1);//左边部分先按x从小到大排序，当x相等时再按y从到大到小排序        for(int i=m/2; i<m; i++)//找右边部分        {            f[i].x=g[i].x;            f[i].y=g[i].y;        }        sort(f+m/2,f+m,cmp2);//右边部分先按x从小到大排序，当x相等时还是按y从到小到大排序        if(m%2)        {            int x=m/2-1;            int y=m/2+1;            int temp=2*f[m/2].x;            for(int i=0,j=m-1; i<m/2; i++,j--)            {                if((f[i].x+f[j].x)!=temp&&f[i].x!=f[j].x)//如果这两点横坐标不对称，并且这里两点的横坐标不相等，图形就肯定是不对称的                {                    ok=1;                    printf("NO\n");                    break;                }                else//横坐标对称（包括横坐标可以相等，也可以不相等）                {                    if(f[i].y==f[j].y||(f[i].x==f[j].x))//y坐标相等或者横坐标相等（因为当横坐标不相等，y坐标相等时，这两点事对称的，这是一种情况，还有就是当横坐标相等时，他也是对称的，因为是平行于y轴啊），下面偶数情况也是一样的分析                        num++;                }            }        }        else        {            int x=m/2-1;            int y=m/2;            int temp=f[x].x+f[y].x;            for(int i=0,j=m-1; i<m/2; i++,j--)            {                if((f[i].x+f[j].x)!=temp&&f[i].x!=f[j].x)                {                    ok=1;                    printf("NO\n");                    break;                }                else                {                    if(f[i].y==f[j].y||(f[i].x==f[j].x))                        num++;                }            }        }        if(!ok)        {            if(num==m/2)                printf("YES\n");            else                printf("NO\n");        }    }    return 0;}