hdu 1071 The area(数学－－抛物线)

最新推荐文章于 2019-06-29 14:23:55 发布

原创最新推荐文章于 2019-06-29 14:23:55 发布 · 652 阅读

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文章标签：

#C++ #数学 #解析几何 #抛物线

本篇介绍了一种通过解析几何方法计算由抛物线和直线围成区域面积的算法实现。利用已知交点坐标，推导出抛物线方程，并通过积分求得准确的面积值。

The area

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8221 Accepted Submission(s): 5772

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

题目分析：可以可以根据抛物线的性质得，x1 = -b/(2*a) ，然后将其余两个点代入，就能求得抛物线的公式和直线公式，进而积分出面积

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

using namespace std;

int t;
double x1 , yy1 , x2 ,y2 , x3 , y3;
double a,b,c,k,d;

int main ( )
{
    scanf ( "%d" , &t );
    while ( t-- )
    {
        scanf ( "%lf%lf%lf%lf%lf%lf" , &x1 , &yy1 , &x2 , &y2 , &x3 , &y3 );
        a = (yy1-y3)/(x1*x1-x3*x3 + 2*x1*x3 - 2*x1*x1 );
        b = -2*x1*a;
        k = (y2-y3)/(x2-x3);
        c = yy1 - a*x1*x1 - b*x1;
        d = y2 - k*x2;
        double A = a/3 , B = (b-k)/2 , C = c - d;
        printf ( "%.2lf\n" , A*(x3*x3*x3-x2*x2*x2) + B*(x3*x3-x2*x2) + C*(x3-x2) );
    }
}