hdu1071-The area 积分求解与x轴形成的面积

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最新推荐文章于 2024-07-29 14:00:00 发布

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分类专栏：杭电随笔计算几何

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本文介绍了一种计算由抛物线和直线围成的土地面积的方法。通过给出三个交点坐标，利用高中数学中的积分知识来求解该土地的具体面积。

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The area

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10810 Accepted Submission(s): 7620

Problem Description

Ignatius bought a land last week, but he didn't know the area of the land because the land is enclosed by a parabola and a straight line. The picture below shows the area. Now given all the intersectant points shows in the picture, can you tell Ignatius the area of the land?

Note: The point P1 in the picture is the vertex of the parabola.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains three intersectant points which shows in the picture, they are given in the order of P1, P2, P3. Each point is described by two floating-point numbers X and Y(0.0<=X,Y<=1000.0).

Output

For each test case, you should output the area of the land, the result should be rounded to 2 decimal places.

Sample Input

  
   2
5.000000 5.000000
0.000000 0.000000
10.000000 0.000000
10.000000 10.000000
1.000000 1.000000
14.000000 8.222222

Sample Output

  
   33.33
40.69

  
  

  
  
   题意：求阴影部分的面积
  
  
   利用高中的积分知识可得：
  
  
   抛物线得到的面积减去直线得到的面积，就是阴影部分的面积
  
  
   求一般方程如下
  
  
   求抛物线方程：
  
  
   已知三点求解抛物线方程：
  
  
   x1^2a+x1b+c=y1
x2^2a+x2b+c=y2
x3^2a+x3b+c=y3
D=x1^2*x2-x1^2*x3-x2^2*x1+x2^2*x3+x3^2*x1-x3^2*x2
D1=y1*x2-y1*x3-y2*x1+y2*x3+y3*x1-y3*x2
D2=x1^2*y2-x1^2*y3-x2^2*y1+x2^2*y3+x3^2*y1-x3^2*y2
D3=x1^2*y3*x2-x1^2*y2*x3-x2^2*y3*x1+x2^2*y1*x3+x3^2*y2*x1-x3^2*y1*x2
A=D1/D
B=D2/D
C=D3/D已知直线求方程：
  
  
   A=(Y1-Y2)/(X1-X2)   B=Y1-A*X1
  
  
   

  
  
   得到参数后直接求解面积就可以了
  
  
   

  
  
   代码如下：
  
  
   #include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
double x1,x2,x3;
double y,y2,y3;
double a,b,c;
double fun(double p,double q)
{
	double A=1.0*a*p*p*p/3+1.0*b*p*p/2+c*p;
	double B=1.0*a*q*q*q/3+1.0*b*q*q/2+c*q;
	return B-A;
}
void init()
{
	double d=x1*x1*x2-x1*x1*x3-x2*x2*x1+x2*x2*x3+x3*x3*x1-x3*x3*x2;
	double d1=y*x2-y*x3-y2*x1+y2*x3+y3*x1-y3*x2;
	double d2=x1*x1*y2-x1*x1*y3-x2*x2*y+x2*x2*y3+x3*x3*y-x3*x3*y2;
	double d3=x1*x1*y3*x2-x1*x1*y2*x3-x2*x2*y3*x1+x2*x2*y*x3+x3*x3*y2*x1-x3*x3*y*x2;
	a=d1/d;
	b=d2/d;
	c=d3/d;
//	printf("a=%.2f\nb=%.2f\nc=%.2f\n",a,b,c);
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf%lf",&x1,&y);
		scanf("%lf%lf",&x2,&y2);
		scanf("%lf%lf",&x3,&y3);
		init();
		double b1=(y2-y3)/(x2-x3);
		double c1=y2-b1*x2;
	//	printf("b1=%.2f\nc1=%.2f\n",b1,c1);
		b=b-b1; c=c-c1;
	//	printf("b=%.2f\nc=%.2f\n",b,c);
		printf("%.2f\n",fun(x2,x3));
	 } 
	return 0;
 } 

   

   点击打开链接http://acm.hdu.edu.cn/showproblem.php?pid=1071