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.
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 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.
OutputPrint exactly one line for each test case. The line should contain 'YES' if the figure is left-right symmetric,and 'NO', otherwise.
Sample Input3 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 14Sample Output
YES NO
YES
这个题意思是给你一些点的坐标,问你这些点是否关于其中一个点对称。做的时候再纠结怎么一次排序才好,看了别人的思路才知道其实排两次就好。
#include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> using namespace std; struct X { int a,b; }; X x1[1003],x2[1003]; bool cmp(X m,X n) { if(m.a==n.a) return m.b>n.b; return m.a<n.a; } bool cmpp(X m,X n) { if(m.a==n.a) return m.b>n.b; return m.a>n.a; } int main() { int t; scanf("%d",&t); while(t--) { memset(x1,0,sizeof(x1)); memset(x2,0,sizeof(x2)); int n,A=10001,B=-10001; scanf("%d",&n); for(int i=1; i<=n; i++) { scanf("%d%d",&x1[i].a,&x1[i].b); x2[i].a=x1[i].a; x2[i].b=x1[i].b; A=min(A,x1[i].a); B=max(B,x1[i].a); } if(A==B) { printf("YES "); continue; } int flag=0; sort(x1+1,x1+n+1,cmp); sort(x2+1,x2+n+1,cmpp); for(int i=1; i<=n; i++) { //cout<<x1[i].a<<x2[i].a<<x1[i].b<<x2[i].b<<endl;; if(x1[i].a+x2[i].a!=(A+B)) { flag=1; break; } if(x1[i].b!=x2[i].b) { flag=1; break; } } if(flag==1) printf("NO "); else printf("YES "); } }