Rectangles

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28381 Accepted Submission(s): 9199

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.

/*在网上看了好多代码，发现好多都是有问题的，
因为好多都忽略了对角线是左上和右下的情况，
这个代码是最好的，它是先计算分别两个矩形
的长和宽的和，再减去非重叠的部分*/ 
#include <bits/stdc++.h>

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]);  //x[3]-x[0]就是非重叠部分，可以自己试一下 
        h = fabs(y2 - y1) + fabs(y4 - y3) - (y[3] - y[0]);  //和上面同理 
        s = l * h;
        if (l <= 0 || h <= 0)
            s = 0.00;
        printf("%.2lf\n", s);
    }
    return 0;

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

hdu 2056 矩形重叠

Rectangles

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