HDU 1071 解方程

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

题意：给你三个点的座标，求面积。

思路：额，没想到好的，强行解三元一次方程组把a,b,c解出来，再利用积分求解。。

这个更简单，抛物线顶点座标表示公式：y=a(x-x1)^2+y1，代入x2或x3座标可解得抛物线公式，再利用x2,x3,解y=kx+b，积分就可以了。

#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<iostream>
#include<cmath>
#include<queue>
#include<map>
using namespace std;

void swap(double *a,double *b){ //遇到除以零的情况时，两行交换位置 
	double t;
	t=*a;
	*a=*b;
	*b=t;
}
int main(){
	freopen("input.txt","r",stdin);
	int t;
	cin>>t;
	while(t--){
		double a,b,c,u,v,uy,vy;
		double x1,y1,x2,y2,x3,y3,x1_2,x2_2,x3_2,m1=1,m2=1,m3=1;
		cin>>x1>>y1>>x2>>y2>>x3>>y3;
		u=x2;uy=y2;
		v=x3;vy=y3;
		
		x1_2=x1*x1;
		x2_2=x2*x2;
		x3_2=x3*x3;
		
		if(x1_2==0){
			swap(&x1_2,&x2_2);
			swap(&x1,&x2);
			swap(&m1,&m2);
			swap(&y1,&y2);
		}
		x2=x2-x2_2/x1_2*x1;
		m2=m2-x2_2/x1_2*m1;
		y2=y2-x2_2/x1_2*y1;
 //       printf("x2=%.2f\nm2=%.2f\ny2=%.2f\n",x2,m2,y2);
        
		if(x1_2==0){
			swap(&x1_2,&x2_2);
			swap(&x3,&x2);
			swap(&m3,&m2);
			swap(&y3,&y2);
		}		
		x3=x3-x3_2/x1_2*x1;
		m3=m3-x3_2/x1_2*m1;
		y3=y3-x3_2/x1_2*y1;
//		printf("x3=%.2f\nm3=%.2f\ny3=%.2f\n",x3,m3,y3);

		if(x2==0){
			swap(&x3,&x2);
			swap(&m3,&m2);
			swap(&y3,&y2);
		}
		m3=m3-x3/x2*m2;
		y3=y3-x3/x2*y2;
//		printf("m3=%.2f\ny3=%.2f\n",m3,y3);
		
		c=y3/m3;
		b=(y2-m2*c)/x2;
		a=(y1-m1*c-x1*b)/x1_2;
		
//		printf("a=%.2f\nb=%.2f\nc=%.2f\n",a,b,c);
		
		double S,S1,S2;
		S1=(a/3*v*v*v+b/2*v*v+c*v)-(a/3*u*u*u+b/2*u*u+c*u);
//		printf("S1=%.2f\n",S1);
		S2=(uy+vy)*(v-u)/2.0;
//		printf("S2=%.2f\n",S2);
		printf("%.2f\n",S1-S2);
	}
	return 0;
}