poj 1701【数学几何】

The area

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6524    Accepted Submission(s): 4578

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1071

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

 

 

Recommend

 

题目描述：给出一个如图所示的抛物线和直线，已知两个交点坐标和抛物线的顶点坐标（顶点坐标是（x1,y1）,两个交点的坐标是（x2,y2）,（x3,y3）），求直线和抛物线所围成的图形的面积。

声明：这道题目并不难，只要学过定积分就一定会做，只不过一次ac了感觉很爽，所以留下此题自勉之~

代码：

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int main()
{
    int zong;
    cin>>zong;
    while(zong--)
    {
        double x1,y1,x2,y2,x3,y3;
        cin>>x1>>y1>>x2>>y2>>x3>>y3;
        double t1=(x2*x2-x3*x3)*(y2-y1)-(x2*x2-x1*x1)*(y2-y3);
        double t2=(x2*x2-x3*x3)*(x2-x1)-(x2*x2-x1*x1)*(x2-x3);
        double b=t1/t2;
        double a=(y2-y1-b*(x2-x1))/(x2*x2-x1*x1);
        double c=y1-a*x1*x1-b*x1;
        double k=(y2-y3)/(x2-x3);
        double d=y2-k*x2;
        double s1=1.0/3.0*a*(x3*x3*x3-x2*x2*x2)+1.0/2*b*(x3*x3-x2*x2)+c*(x3-x2);
        double s2=(k*(x2+x3)+2*d)*(x3-x2)*1.0/2.0;
        double s=s1-s2;
        printf("%.2lf\n",s);
    }
    return 0;
}