Hdu 1071 The area 利用计算几何求面积

The area

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7793    Accepted Submission(s): 5472

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.
 

 

Author

Ignatius.L

这一题是关于抛物线与直线相交，求围成的阴影部分的面积。

已知抛物线顶点P1坐标（-b/(2*a),(4*a*c-b*b)/(4*a)），

抛物线公式y=ax^2+bx+c;

直线的一般式y=kx+h;

由三个抛物线上的三个点可以求出抛物线方程

y1= ax1^2+bx1+c

y2= ax2^2+bx2+c

又x1=-b/(2a)

所以可得

 a=(y2-y1)/(x2-x1)^2

 b=-2ax1

 c=y1-ax1^2-bx1

由y2=kx2+h

    y3=kx3+h

所以

k=(y3-y2)/(x3-x2)

        h=y2-kx2

面积公式S=a/3*(x3^3-x2^3)+(b-k)/2*(x3^2-x2^2)+(c-h)*(x3-x2)

#include <iostream>
#include <cstdio>
using namespace std;
struct Point {
    double x,y;
}P[3];
int main()
{
    int T;
    double a, b, c, k, h, S;
    scanf("%d",&T);
    while(T--)
    {
        S = 0;
        scanf("%lf%lf",&P[0].x,&P[0].y);
        scanf("%lf%lf",&P[1].x,&P[1].y);
        scanf("%lf%lf",&P[2].x,&P[2].y);
        a = (P[1].y-P[0].y)/((P[1].x-P[0].x)*(P[1].x-P[0].x));
        b = -2*a*P[0].x;
        c = P[0].y-a*P[0].x*P[0].x-b*P[0].x;
        k = (P[2].y-P[1].y)/(P[2].x-P[1].x);
        h = P[1].y-k*P[1].x;
        S = a/3*(P[2].x*P[2].x*P[2].x-P[1].x*P[1].x*P[1].x)+(b-k)/2*(P[2].x*P[2].x-P[1].x*P[1].x)+(c-h)*(P[2].x-P[1].x);
        printf("%.2lf\n",S);
    }
    return 0;
}