HDU 5120 Intersection

標籤：

Intersection

Problem DescriptionMatt 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. 

 

InputThe 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. 

 

OutputFor 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 Input22 30 00 02 30 05 0  

 

Sample OutputCase #1: 15.707963Case #2: 2.250778

#include<iostream>#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef double db;const db PI=acos(-1.0);struct point{    db x,y;}p[2];struct circle{    point O;    db r;}C[4];db distcal(point a,point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}db areacal(circle a,circle b){    if(a.r>b.r)swap(a,b);    db d=distcal(a.O,b.O);    if(d==0.0||d+a.r<=b.r){db t=a.r;return PI*t*t;}    if(a.r+b.r<=d)return 0.0;    db th1=acos((a.r*a.r+d*d-b.r*b.r)/(2*a.r*d));    db th2=acos((b.r*b.r+d*d-a.r*a.r)/(2*b.r*d));    db are1=a.r*a.r*th1;    db are2=b.r*b.r*th2;    db are3=a.r*d*sin(th1);    return are1+are2-are3;}int main(){    int T;    scanf("%d",&T);    for(int kase=1;kase<=T;kase++)    {        db r,R;        scanf("%lf%lf",&r,&R);        scanf("%lf%lf%lf%lf",&p[0].x,&p[0].y,&p[1].x,&p[1].y);        C[0].O=p[0],C[0].r=R;        C[1].O=p[0],C[1].r=r;        C[2].O=p[1],C[2].r=R;        C[3].O=p[1],C[3].r=r;        db ans=areacal(C[0],C[2])+areacal(C[1],C[3])-2.0*areacal(C[0],C[3]);        printf("Case #%d: %.6f\n",kase,ans);    }    return 0;}/*11 100 01 0*/

 

 

HDU 5120 Intersection