HDOJ 1071(球泡无线和直线区域内的面积)

#include <stdio.h>   
typedef struct
{   
    double x,y;   
}Point;   
    
double area(Point p1, Point p2, Point p3)
{   
    double a, b, c;   // 抛物线的参数   
    double k, h;      // 直线的参数   
    double s;   
    // 抛物线 y = ax^2 + bx + c   
    a = (p2.y - p1.y) / ((p2.x - p1.x)*(p2.x - p1.x));   // a = (y2-y1) / (x2-x1)^2   
    b = -2 * a * p1.x;    // b = -2a*x1;   
    //c = p1.y - a * p1.x * p1.x - b * p1.x;   // c = y - ax^2 - bx   
    c=a*p1.x*p1.x+p1.y;
    //直线 y = kx + h   
    k = (p2.y - p3.y) / (p2.x - p3.x);   
    h = p3.y - k * p3.x;   
  
    s = a/3*(p3.x*p3.x*p3.x - p2.x*p2.x*p2.x) + (b-k)/2*(p3.x*p3.x-p2.x*p2.x) + (c-h)*(p3.x-p2.x);   
    return s;   
}   
  
int main(){   
  
    Point p1, p2, p3;   
    int T;   
    scanf("%d", &T);   
    while(T--)   
    {   
        scanf("%lf %lf", &p1.x, &p1.y);   
        scanf("%lf %lf", &p2.x, &p2.y);   
        scanf("%lf %lf", &p3.x, &p3.y);   
    
        printf("%.2lf\n", area(p1,p2,p3));   
    }   
    return 0;      
}
/*
注意:p1是抛物线顶点，利用  y= a(x-x1)^2+y1,求出a，再利用 b/(-2a)=x1,求出b，
实际上根据最值公式求出  c=a*x1^2+y1 
*/