*寒假水19——Rectangles （矩形重叠求重叠面积）

最新推荐文章于 2021-05-19 13:59:42 发布

原创最新推荐文章于 2021-05-19 13:59:42 发布 · 348 阅读

1 ·

CC 4.0 BY-SA版权

2017寒假专栏收录该内容

123 篇文章

订阅专栏

本文介绍了一种计算两个矩形交集面积的方法，输入为两对矩形对角线顶点坐标，输出精确到小数点后两位的交集面积。

Given two rectangles and the coordinates of two points on the diagonals of each rectangle,you have to calculate the area of the intersected part of two rectangles. its sides are parallel to OX and OY .

InputInput The first line of input is 8 positive numbers which indicate the coordinates of four points that must be on each diagonal.The 8 numbers are x1,y1,x2,y2,x3,y3,x4,y4.That means the two points on the first rectangle are(x1,y1),(x2,y2);the other two points on the second rectangle are (x3,y3),(x4,y4).OutputOutput For each case output the area of their intersected part in a single line.accurate up to 2 decimal places.Sample Input

1.00 1.00 3.00 3.00 2.00 2.00 4.00 4.00
5.00 5.00 13.00 13.00 4.00 4.00 12.50 12.50

Sample Output

1.00
56.25

#include<stdio.h>

double max(double a,double b);
double min(double m,double n);

int main()
{
	double x1,y1,x2,y2,x3,y3,x4,y4,x,y,tmp;
	while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4)!=EOF)
	{
		if(x1>x2) tmp=x1,x1=x2,x2=tmp;
		if(x3>x4) tmp=x3,x3=x4,x4=tmp;
        x=min(x2,x4)-max(x1,x3);
    	if(y1>y2) tmp=y1,y1=y2,y2=tmp;
		if(y3>y4) tmp=y3,y3=y4,y4=tmp;
        y=min(y2,y4)-max(y1,y3);
		printf("%.2lf\n",x<0||y<0?0:x*y);		
	}
	return 0;
}

double max(double a,double b)
{
	if(a>b) return a;
	else return b;
}

double min(double m,double n)
{
	if(m>n) return n;
	else return m;
}

题解：暴力算就好。