fzu 2035 Axial symmetry(枚举+几何）

旧城等待， 2021-11-19 13:30 158阅读 0赞

[题目链接：fzu 2035 Axial symmetry][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;
    }

转载于:https://www.cnblogs.com/riasky/p/3471425.html

[fzu 2035 Axial symmetry]: http://acm.fzu.edu.cn/problem.php?pid=2035