fzu 2035 Axial symmetry(枚举+几何）

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题目链接：fzu 2035 Axial symmetry

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;}

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