HDoj 2056 Rectangles

Problem Description

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 .

Input

Input 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).

Output

Output 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

Author

seeyou

Source

校庆杯Warm Up

Recommend

linle   |   We have carefully selected several similar problems for you:  2057 2058 2062 2060 2059

注意两个矩形完全不相交时不能使用公式计算，直接为0

C++代码如下：

#include <iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
    double x[4];
    double y[4];
    while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x[0],&y[0],&x[1],&y[1],&x[2],&y[2],&x[3],&y[3]) !=EOF )
    {
        if( (min(x[0],x[1])>max(x[2],x[3]))  ||  (max(x[0],x[1])<min(x[2],x[3])) )         //首先排除两个矩形左右错开，此时面积为0
        {
            printf("0.00\n");
            continue;
        }
        if( (min(y[0],y[1])>max(y[2],y[3]))  ||  (max(y[0],y[1])<min(y[2],y[3])) )         //排除两个矩形上下错开，此时面积为0
        {
            printf("0.00\n");
            continue;
        }
        sort(x,x+4);
        sort(y,y+4);
        printf("%.2lf\n",(x[2]-x[1])*(y[2]-y[1]));
    }
}

原文地址：https://www.cnblogs.com/wzmm/p/12704621.html