HDU 5120 Intersection （计算几何）

题意:

求两个一模一样,但位置不一定相同的圆环的环相交面积

分析:

其实就是圆交面积++−−

ans=大圆交−2∗一大一小圆交+小圆交

写圆交函数的时候,注意有三种情况,相含,相交,相离

把重合当成相含了, 一直没考虑相含卡到死。。。

代码:

//
//  Created by TaoSama on 2015-10-11
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double PI = acos(-1), EPS = 1e-10;

int sgn(double a) {
return a < -EPS ? -1 : a < EPS ? 0 : 1;
}

struct Point {
double x, y;
};

double getDis(const Point& p1, const Point& p2) {
return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));
}

double getAngle(double a, double b, double c) {
return acos((a * a + b * b - c * c) / 2 / a / b);
}

struct Circle {
double r;
Point p;
Circle() {}
Circle(double r, double x, double y) {
this->r = r;
p = (Point) {x, y};
}
};

double getIntersectionArea(const Circle& c1, const Circle& c2) {
double dis = getDis(c1.p, c2.p);
if(sgn(dis - (c1.r + c2.r)) > 0) return 0;
if(sgn(dis - abs(c1.r - c2.r)) <= 0) {
double minr = min(c1.r, c2.r);
return PI * minr * minr;
}
double alpha = getAngle(c1.r, dis, c2.r) * 2;
double beta = getAngle(c2.r, dis, c1.r) * 2;
double area = 0.5 * alpha * c1.r * c1.r + 0.5 * beta * c2.r * c2.r;
area -= 0.5 * c1.r * c1.r * sin(alpha) + 0.5 * c2.r * c2.r * sin(beta);
return area;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
int kase = 0;
while(t--) {
double r, R;
double x1, y1, x2, y2;
scanf("%lf%lf%lf%lf%lf%lf", &r, &R, &x1, &y1, &x2, &y2);
Circle a1(R, x1, y1), a2(R, x2, y2);
Circle b1(r, x1, y1), b2(r, x2, y2);
double ans = getIntersectionArea(a1, a2) - getIntersectionArea(a1, b2)
- getIntersectionArea(a2, b1) + getIntersectionArea(b1, b2);
printf("Case #%d: %.6f\n", ++kase, ans);
}
return 0;
}