Hdu 5120 Intersection【計算圓環相交面積】
阿新 • • 發佈:2019-02-12
Intersection
Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 1861 Accepted Submission(s): 709Problem 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
Each of the following two lines contains two integers xi, yi (0 ≤ xi, yi ≤ 20) indicating the coordinates of the center of each ring.
Output For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the area of intersection rounded to 6 decimal places.
Sample Input 2 2 3 0 0 0 0 2 3 0 0 5 0
Sample Output Case #1: 15.707963 Case #2: 2.250778
Source
題意:
給出兩個相同的圓環的內徑和外徑,以及圓心的座標,求兩個圓環交叉部分的面積
題解:
前幾天周練,數學大神1A此題,賽後,個人進行嘗試,測試樣例都過不去......真心比不起啊......
一個函式,引數傳遞是兩個圓的半徑,以及兩個圓的圓心座標,返回兩個圓的交叉部分面積
然後就是容斥原理的一部分了:
S=S(兩個外環大圓相交部分)-2* S(一個的外環和另一個的內環的相交面積)+S(兩個內環的圓的相交部分)
好久沒接觸過幾何題,感覺自己好水.......
/*
http://blog.csdn.net/liuke19950717
*/
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double pi=acos(-1.0);
struct node
{
double x,y;
};
double dis(node a,node b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double circle_cross(double r1,double r2,node a,node b)
{
double d=dis(a,b);//兩圓的間距
if(r1<r2)//保證大圓在前
{
swap(r1,r2);
}
if(d>=r1+r2)
{
return 0;
}
if(d<=r1-r2)
{
return pi*r2*r2;
}
double deg1=acos((r1*r1+d*d-r2*r2)/(2*r1*d));
double deg2=acos((r2*r2+d*d-r1*r1)/(2*r2*d));
return deg1*r1*r1+deg2*r2*r2-r1*sin(deg1)*d;
}
int main()
{
int t;
scanf("%d",&t);
for(int k=1;k<=t;++k)
{
int r,R;
node p[5];
scanf("%d%d",&r,&R);
for(int i=1;i<=2;++i)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
}
double d=dis(p[1],p[2]);
double a=circle_cross(R,R,p[1],p[2]);
double b=circle_cross(R,r,p[1],p[2]);
double c=circle_cross(r,r,p[1],p[2]);
double ans=a+c-2*b;
printf("Case #%d: %.6f\n",k,ans);
}
return 0;
}