hdu 1071 - The area

The area

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6004    Accepted Submission(s): 4193

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.

====================================

直线方程：y=kx+b

抛物线方程：一般式：y=ax^2+bx+c

                        顶点式：y=a(x-h)^2+c

两方程相减做积分

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

int main()
{
    int t;double x0,x1,x2,y0,y1,y2,k,b,h,c,a,s;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf%lf%lf",&x0,&y0,&x1,&y1,&x2,&y2);
        k=(y2-y1)/(x2-x1);
        b=y1-k*x1;
        h=x0;
        c=y0;
        a=(y1-c)/((x1-h)*(x1-h));
        s=a/3*((x2-h)*(x2-h)*(x2-h)-(x1-h)*(x1-h)*(x1-h))+c*(x2-x1)-k/2*(x2*x2-x1*x1)-b*(x2-x1);
        printf("%.2lf\n",s);
    }
    return 0;
}