HDU 1071 The area(抛物线与直线围成的面积）

The area

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10809    Accepted Submission(s): 7619

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

Hint

For float may be not accurate enough, please use double instead of float.

 

题目大意：给你三个点，P1是抛物线的顶点，P2和P3是抛物线和直线的交点，求抛物线和直线围成的面积。

抛物线方程为 x^2a+xb+c=y

那么

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

对于直线方程 y=ax+b;

a=(y3-y2)/(x3-x2);

b=y2-a*x2;

如上直接求即可

如下是AC代码

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		double x1,x2,x3,y1,y2,y3;
		scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
		double a,b,c,a1,b1;
		double d=x1*x1*x2-x1*x1*x3-x2*x2*x1+x2*x2*x3+x3*x3*x1-x3*x3*x2;
		double d1=y1*x2-y1*x3-y2*x1+y2*x3+y3*x1-y3*x2;
		double d2=x1*x1*y2-x1*x1*y3-x2*x2*y1+x2*x2*y3+x3*x3*y1-x3*x3*y2;
		double d3=x1*x1*y3*x2-x1*x1*y2*x3-x2*x2*y3*x1+x2*x2*y1*x3+x3*x3*y2*x1-x3*x3*y1*x2;
		a=d1/d;
		b=d2/d;
		c=d3/d;
		a1=(y3-y2)/(x3-x2);
		b1=y2-a1*x2;
		double ans=a*x3*x3*x3/3.0+((b-a1)/2.0)*x3*x3+(c-b1)*x3;
		ans-=a*x2*x2*x2/3.0+((b-a1)/2.0)*x2*x2+(c-b1)*x2;
		printf("%.2f\n",ans);
	}
	return 0;
}