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lightoj1118 Incredible Molecules

2015-12-02 20:01 423 查看
思路:求两个圆相互遮盖的面积,对于圆的位置情况有外切,内切,相离,内含,相交。

这里需要考虑的是(相离+外切),相交,内含(两种情况);

手推了下公式,,刚入门,,,推得有点慢。

// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
// #define DEBUG
#ifdef DEBUG
#define debug(...) printf( __VA_ARGS__ )
#else
#define debug(...)
#endif
#define CLR(x) memset(x, 0,sizeof x)
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
template<class T> inline T Get_Max(const T&a,const T&b){return a < b?b:a;}
template<class T> inline T Get_Min(const T&a,const T&b){return a < b?a:b;}
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const double eps = 1e-10;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Point{
double x, y;
Point(){}
Point(double _x, double _y){
x = _x;
y = _y;
}
Point operator + (const Point& rhs)const{
return Point(x + rhs.x, y + rhs.y);
}
Point operator - (const Point& rhs)const{
return Point(x - rhs.x, y - rhs.y);
}
double operator ^ (const Point& rhs)const{
return (x * rhs.y - y * rhs.x);
}
double operator * (const Point& rhs)const{
return (x * rhs.x + y * rhs.y);
}
Point operator * (const double Num)const{
return Point(x * Num,y * Num);
}
friend ostream& operator << (ostream& output,const Point& rhs){
output << "(" << rhs.x << "," << rhs.y << ")";
return output;
}
};
typedef Point Vector;
/*向量的模*/
inline double Length(const Vector& A){//Get the length of vector A;
return sqrt(A * A);//sqrt(x*x+y*y);
}
/*向量夹角*/
inline double Angle(const Vector& A,const Point& B){
return acos(A * B/Length(A)/Length(B));
}
/*向量A旋转rad弧度*/
inline Point Rotate(const Vector& A, double rad){
return Vector(A.x*cos(rad) - A.y*sin(rad),A.x*sin(rad) + A.y*cos(rad));
}
double mul(double x){
return x*x;
}
Point A,B;
double R,r;
const double PI = acos(-1.0);
int main()
{	// ios::sync_with_stdio(false);
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int t, icase = 0;
cin >> t;
while(t--){
cin >> A.x >> A.y >> R;
cin >> B.x >> B.y >> r;
double l = Length(A - B);
if (l >= R + r){
printf("Case %d: 0\n", ++icase);
continue;
}//像离
if (l + r <= R && R >= r){
// debug("Cao\n");
printf("Case %d: %.6lf\n", ++icase, PI*r*r);
continue;
}//内含
if (l + R <= r && r >= R){
// debug("here\n");
printf("Case %d: %.6lf\n", ++icase, PI*R*R);
continue;
}//内含
double a1 = acos((mul(R) + mul(l) - mul(r))/2.0/R/l);
double a2 = acos((mul(l) + mul(r) - mul(R))/2.0/r/l);
double s = a1*R*R + a2*r*r - (R*R*sin(2*a1)/2.0 + r*r*sin(2*a2)/2.0);
printf("Case %d: %.6lf\n", ++icase, s);
}
return 0;
}
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