首页 > ACM题库 > HDU-杭电 > HDU 1071 The area-计算几何-[解题报告] C++
2013
11-26

HDU 1071 The area-计算几何-[解题报告] C++

The area

问题描述 :

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.

输入:

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).

输出:

For each test case, you should output the area of the land, the result should be rounded to 2 decimal places.

样例输入:

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

样例输出:

33.33
40.69

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

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

推定积分公式。。。。

设直线方程:y=kx+t…………………………………………………………(1)
  抛物线方程:y=ax^2+bx+c……………………………………………………(2)
已知抛物线顶点p1(x1,y1),两线交点p2(x2,y2)和p3(x3,y3)
斜率k=(y3-y2)/(x3-x2)……………………………………………………(3)
把p3点代入(1)式结合(3)式可得:t=y3-(k*x3)
又因为p1是抛物线的顶点,可得关系:x1=-b/2a即b=-2a*x1………………(4)
把p1点代入(2)式结合(4)式可得:a*x1*x1-2a*x1*x1+c=y1化简得c=y1+a*x1*x1……(5)
把p2点代入(2)式结合(4)式和(5)式可得:a=(y2-y1)/((x1-x2)*(x1-x2))
于是通过3点求出了k,t,a,b,c即两个方程式已求出
题目时求面积s
通过积分可知:s=f(x2->x3)(积分符号)(ax^2+bx+c-(kx+t))
               =f(x2->x3)(积分符号)(ax^2+(b-k)x+c-t)
               =[a/3*x^3+(b-k)/2*x^2+(c-t)x](x2->x3)
               =a/3*x3*x3*x3+(b-k)/2*x3*x3+(c-t)*x3-(a/3*x2*x2*x2+(b-k)/2*x2*x2+(c-t)*x2)
化简得:
面积公式:s=-(y2-y1)/((x2-x1)*(x2-x1))*((x3-x2)*(x3-x2)*(x3-x2))/6;

#include<stdio.h>
int main()
{
    double a,b,c,t,k,x1,x2,x3,y1,y2,y3,s1,s2;
    int te;
    scanf("%d",&te);
    while(te--)
    {
       scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
       k=(y3-y2)/(x3-x2);
       t=y3-k*x3;
       a=(y2-y1)/((x1-x2)*(x1-x2));
       b=-x1*2*a;
       c=y1-a*x1*x1-b*x1;
       s1=1.0/3*a*x2*x2*x2+1.0/2*(b-k)*x2*x2+x2*(c-t);
       s2=1.0/3*a*x3*x3*x3+1.0/2*(b-k)*x3*x3+x3*(c-t);
       printf("%.2lf\n",s2-s1);
    }
    return 0;
}

 


  1. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。