Southern Subregion Problem C. Equivalent Cards

最新推荐文章于 2021-05-18 01:03:29 发布

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本文介绍了一个基于几何计算的程序，使用C++实现。该程序通过定义点结构体进行坐标输入，并计算两点间的距离及向量叉乘等操作。通过对输入的多个四边形面积求和并与两个固定圆的交集面积比较，来评估不同区域的相似性。最终通过并查集的方法确定这些区域是否属于同一类别。

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
#define maxn 105
using namespace std;
int n;
double val[maxn];
struct Point{
		double x,y;
		Point(double x=0.0,double y = 0.0):x(x),y(y){}
		void init(){
				cin >> x >> y;
		}
		friend Point operator - (const Point &a, const Point &b){
				return Point(a.x - b.x,a.y - b.y);
		}
};
double sum1;
double sum2;
Point point1[2],point2[2];
double abss(double a){
		return a < 0 ? -a : a;
}
const double pi = 3.1415926535;
double cross(const Point &a, const Point &b){
		return a.x * b.y - a.y * b.x;
}
double dist(const Point &a, const Point &b){
		double dx = a.x - b.x;
		double dy = a.y - b.y;
		return sqrt(dx*dx + dy*dy);
}
void deal(int x){
		sum1 = 0;
		sum2 = 0;
		for(int i = 0; i < 2; i++) point1[i].init();
		for(int i = 0; i < 2; i++) point2[i].init();
		double r = dist(point1[0],point1[1]) / 2;
		double r2 = dist(point2[0],point2[1]) / 2;
		sum1 = pi * r * r2;
		int k; cin >> k;
		for(int i = 1; i <= k; i++) {
				Point a,b,c,d;
				a.init();
				b.init();
				c.init();
				d.init();
				sum2 += abss(cross(b-a,c-a));
				sum2 += abss(cross(c-a,d-a));
		}
		val[x] = sum2/sum1;
}
const double expp = 1e-5;
int sgn(double a){
		if(abss(a) < expp) return 0;
		else return a < 0 ? -1 : 1;
}
int fa[maxn];
int find(int x){
		if(fa[x] == x) return x;
		else return fa[x] = find(fa[x]);
}
void work(){
		for(int i = 1; i <= n; i++) fa[i] = i;
		for(int i = 1; i <= n; i++) {
				for(int j = i+1; j <= n; j++) {
						int fx = find(i);
						int fy = find(j);
						if(sgn(val[i] - val[j]) == 0) {
								fa[fy] = fx;
						}
				}
		}	
		/*for(int i = 1; i <= n; i++) {
				cout << val[i] << " ";
		}
		cout << endl;*/
		for(int i = 1; i <= n; i++) {
				cout << find(i) << " ";
		}
		cout << endl;
}
int main(){
		//freopen("main.in","r",stdin);
		ios::sync_with_stdio(false);
		cin >> n;
		for(int i = 1; i <= n; i++){
				deal(i);
		}
		work();
		return 0;
}