HDU 2056 Rectangles

浏览: 26 发布日期: 2017-12-29 分类: c

Rectangles

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 26882 Accepted Submission(s): 8705

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

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

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


http://acm.hdu.edu.cn/showpro...

Accepted Code

#include <iostream>
#include <cmath>
#include <algorithm>

using namespace std;

int main()
{
    double x1, y1, x2, y2, x3, y3, x4, y4;
    double x[4], y[4];
    double s, l, h;

    while (cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3 >> x4 >> y4)
    {
        x[0] = x1; x[1] = x2; x[2] = x3; x[3] = x4;
        y[0] = y1; y[1] = y2; y[2] = y3; y[3] = y4;

        sort(x, x + 4);
        sort(y, y + 4);

        l = fabs(x2 - x1) + fabs(x4 - x3) - (x[3] - x[0]);
        h = fabs(y2 - y1) + fabs(y4 - y3) - (y[3] - y[0]);

        s = l * h;

        if (l <= 0 || h <= 0)
            s = 0.00;

        printf("%.2lf\n", s);
    }

    return 0;
}

Notes

题意:
给四个平面直角坐标,可以构成两个矩形,求相交面积。

感受:
直接暴力求解很容易疏忽,改进后的求长宽的方法(见代码)就有效地避免了考虑疏忽。

//-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
//
// 作者:龙威昊
// 完成时间:2017/12/27
//
//-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

返回顶部