hdu 5120 Intersection 圆环面积交

Intersection

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5120

Description

Matt is a big fan of logo design. Recently he falls in love with logo made up by rings. The following figures are some famous examples you may know.

A ring is a 2-D figure bounded by two circles sharing the common center. The radius for these circles are denoted by r and R (r < R). For more details, refer to the gray part in the illustration below.

Matt just designed a new logo consisting of two rings with the same size in the 2-D plane. For his interests, Matt would like to know the area of the intersection of these two rings.

Input

The first line contains only one integer T (T ≤ 105), which indicates the number of test cases. For each test case, the first line contains two integers r, R (0 ≤ r < R ≤ 10).

Each of the following two lines contains two integers xi, yi (0 ≤ xi, yi ≤ 20) indicating the coordinates of the center of each ring.

HINT

题意

给你两个一样的圆环，问你这两个圆环相交的面积是多少

题解：

容斥定理就好了，剩下的都是套版

面积=大交大-大交小+小交小

代码：

//qscqesze
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1000006
#define mod 1001
#define eps 1e-9
#define PI acos(-1)
const double EP  = 1E-10 ;
int Num;
//const int inf=0x7fffffff;
const ll inf=999999999;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
//*************************************************************************************
struct Circle{
    double x,y;
    double r;
};

double calArea(Circle c1, Circle c2)
{
    double d;
    double s,s1,s2,s3,angle1,angle2,temp;

    d=sqrt((c1.x-c2.x)*(c1.x-c2.x)+(c1.y-c2.y)*(c1.y-c2.y));
    if(d>=(c1.r+c2.r))//两圆相离
        return 0;
    if((c1.r-c2.r)>=d)//两圆内含,c1大
        return acos(-1.0)*c2.r*c2.r;
    if((c2.r-c1.r)>=d)//两圆内含,c2大
        return acos(-1.0)*c1.r*c1.r;

    angle1=acos((c1.r*c1.r+d*d-c2.r*c2.r)/(2*c1.r*d));
    angle2=acos((c2.r*c2.r+d*d-c1.r*c1.r)/(2*c2.r*d));

    s1=angle1*c1.r*c1.r;s2=angle2*c2.r*c2.r;
    s3=c1.r*d*sin(angle1);
    s=s1+s2-s3;

    return s;
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int cas = 1;cas <= t;cas++)
    {
        Circle a1,a2,b1,b2;
        double R,r;
        scanf("%lf%lf",&a1.r,&a2.r);
        b1.r=a1.r;
        b2.r=a2.r;
        scanf("%lf%lf",&a1.x,&a1.y);
        scanf("%lf%lf",&b1.x,&b1.y);
        a2.x=a1.x;
        a2.y=a1.y;
        b2.x=b1.x;
        b2.y=b1.y;
        double ans = 0;
        ans += calArea(a1,b1);
        ans -= calArea(a1,b2);
        ans -= calArea(b1,a2);
        ans += calArea(a2,b2);
        printf("Case #%d: %.6lf\n",cas,ans);
    }
}

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原文地址：http://www.cnblogs.com/qscqesze/p/4817578.html