轉載

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兩圓相交分如下集中情況：相離、相切、相交、包含。

設兩圓圓心分別是O1和O2，半徑分別是r1和r2，設d為兩圓心距離。又因為兩圓有大有小，我們設較小的圓是O1。

相離相切的面積為零，代碼如下：

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[cpp] view plaincopy print?

double?d?=?sqrt((a.x-b.x)*(a.x-b.x)?+?(a.y-b.y)*(a.y-b.y));??

if?(d?>=?r1+r2)??

????return?0;??

double d = sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y)); if (d >= r1+r2)return 0;

包含的面積就是小圓的面積了，代碼如下：

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[cpp] view plaincopy print?

if(r2?-?r1?>=?d)??

????return?pi*r1*r1;??

if(r2 - r1 >= d)return pi*r1*r1;

接下來看看相交的情況。

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相交面積可以這樣算：扇形O1AB - △O1AB + 扇形O2AB - △O2AB，這兩個三角形組成了一個四邊形，可以用兩倍的△O1AO2求得，

所以答案就是兩個扇形-兩倍的△O1AO2

因為

所以

那么

同理

接下來是四邊形面積：

代碼如下：

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double ang1=acos((r1*r1+d*d-r2*r2)/(2*r1*d)); double ang2=acos((r2*r2+d*d-r1*r1)/(2*r2*d)); return ang1*r1*r1 + ang2*r2*r2 - r1*d*sin(ang1);

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#include<iostream> #include<cmath> using namespace std;#define pi acos(-1.0)typedef struct node {int x;int y; }point;double AREA(point a, double r1, point b, double r2) {double d = sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));if (d >= r1+r2)return 0;if (r1>r2){double tmp = r1;r1 = r2;r2 = tmp;}if(r2 - r1 >= d)return pi*r1*r1;double ang1=acos((r1*r1+d*d-r2*r2)/(2*r1*d));double ang2=acos((r2*r2+d*d-r1*r1)/(2*r2*d));return ang1*r1*r1 + ang2*r2*r2 - r1*d*sin(ang1); }int main() {point a, b;a.x=2, a.y=2;b.x=7, b.y=2;double result = AREA(a, 3, b, 5);printf("%lf\n", result);return 0; }

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Intersection

Time Limit: 4000/4000 MS (Java/Others)????Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 3443????Accepted Submission(s): 1302

Problem 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.

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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 x_i, y_i (0 ≤ x_i, y_i ≤ 20) indicating the coordinates of the center of each ring.

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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.

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Sample Input 2 2 3 0 0 0 0 2 3 0 0 5 0

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Sample Output Case #1: 15.707963 Case #2: 2.250778 #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; double x1,ya,x2,y2,dis,s1,s2,s3,R,r; double sov(double R,double r){if(dis>=r+R) return 0;if(dis<=R-r) return acos(-1.0)*r*r;double x=(R*R-r*r+dis*dis)/2.0/dis;double y=(r*r-R*R+dis*dis)/2.0/dis;double seta1=2*acos(x/R);double seta2=2*acos(y/r);double ans=seta1*R*R/2.0+seta2*r*r/2.0;double h=sqrt(R*R-x*x);return ans-dis*h; } int main(){int tas=1,T;for(scanf("%d",&T);T--;){scanf("%lf%lf",&r,&R);scanf("%lf%lf%lf%lf",&x1,&ya,&x2,&y2);dis=sqrt((x1-x2)*(x1-x2)+(ya-y2)*(ya-y2));s1=sov(R,R),s2=sov(R,r),s3=sov(r,r);printf("Case #%d: %.6f\n",tas++,s1-2*s2+s3);} }

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轉載于:https://www.cnblogs.com/mfys/p/7622133.html

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