HDU 5120 Intersection (圆的面积交)

题意：给定两个圆环，求两个圆环的面积交。

析：很容易知道，圆环面积交就是，大圆与大圆面积交 - 大圆和小圆面积交 - 小圆和大圆面积交 + 小圆和小圆面积交。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 5;
const int mod = 1000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r > 0 && r <= n && c > 0 && c <= m;
}

int cmp(double x){
  if(fabs(x) < eps)  return 0;
  return x > 0 ? 1 :-1;
}
struct Circle  {
  double x, y;
  double r;
};
Circle big1, big2, small1, small2;

double dis(const Circle &a, const Circle &b){
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

double solve(const Circle &a, const Circle &b){
  double d = dis(a,b);
  if(d >= a.r + b.r) return 0;
  if(d <= fabs(a.r - b.r)){
    double r = a.r < b.r ? a.r : b.r;
    return PI * r * r;
  }
  double ang1 = acos((a.r * a.r + d * d - b.r * b.r)/2. / a.r / d);
  double ang2 = acos((b.r * b.r + d * d - a.r * a.r)/2. / b.r / d);
  double ret = ang1 * a.r * a.r + ang2 * b.r * b.r - d * a.r * sin(ang1);
  return ret;
}

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%lf %lf", &small1.r, &big1.r);
    small2.r = small1.r;
    big2.r = big1.r;
    scanf("%lf %lf", &small1.x, &small1.y);
    scanf("%lf %lf", &small2.x, &small2.y);
    big1.x = small1.x;
    big1.y = small1.y;
    big2.x = small2.x;
    big2.y = small2.y;
    double ans = solve(big1, big2) - solve(big1, small2) - solve(big2, small1) + solve(small1, small2);
    printf("Case #%d: %.6f\n", kase, ans);
  }
  return 0;
}