fzu 2035 Axial symmetry(枚举+几何）
题目大意：给出n个点，表示n边形的n个顶点，判断该n边形是否为轴对称图形。（给出点按照图形的顺时针或逆时针给出。
解题思路：将相邻两个点的中点有序的加入点集，即保证点是按照图形的顺时针或逆时针出现的，然后枚举i和i + n两点的直线作为对称轴。判断其他所有点是否对称即可。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
const int N = 1005;
const double eps = 1e-9;
struct point {
double x, y;
void get() { scanf("%lf%lf", &x, &y); }
void set(double a, double b) { x = a, y = b; }
bool operator == (const point& c) {
return fabs(x - c.x) < eps && fabs(y - c.y) < eps;
}
}p[N];
int n, vis[N];
void init() {
scanf("%d", &n);
p[0].get();
int t;
for (int i = 1; i < n; i++) {
t = i * 2;
p[t].get();
p[t - 1].set((p[t - 2].x + p[t].x) / 2, (p[t - 2].y + p[t].y) / 2);
}
t = (n - 1) * 2;
p[t + 1].set((p[t].x + p[0].x) / 2, (p[t].y + p[0].y) / 2);
}
void findLine(double& A, double& B, double& C, const point& u, const point& v) {
A = v.y - u.y;
B = u.x - v.x;
C = u.y * (v.x - u.x) + (u.y - v.y) * u.x;
}
point findPoint(double A, double B, double C, point k) {
point c;
c.x = ( (B * B - A * A) * k.x - 2 * A * B * k.y - 2 * A * C ) / (A * A + B * B);
c.y = (-2 * A * B * k.x + (A * A - B * B)* k.y - 2 * B * C) / (A * A + B * B);
return c;
}
bool search(point k) {
for (int i = 0; i < n; i++) {
if (vis[i * 2]) continue;
if (p[i * 2] == k) {
vis[i * 2] = 1;
return true;
}
}
return false;
}
bool judge(double A, double B, double C) {
for (int i = 0; i < n; i++) {
if (vis[i * 2]) continue;
point k = findPoint(A, B, C, p[i * 2]);
if (!search(k)) return false;
}
return true;
}
bool solve() {
double A, B, C;
for (int i = 0; i < n; i++) {
findLine(A, B, C, p[i], p[i + n]);
memset(vis, 0, sizeof(vis));
vis[i] = vis[i + n] = 1;
if (judge(A, B, C) ) return true;
}
return false;
}
int main () {
int cas;
scanf("%d", &cas);
for (int i = 1; i <= cas; i++) {
init();
printf("Case %d: %s\n", i, solve() ? "YES" : "NO");
}
return 0;
}