杭电2056之Rectangles 杭电

2019-11-06 07:16:28

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PRoblem DescriptionGiven 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 4.005.00 5.00 13.00 13.00 4.00 4.00 12.50 12.50Sample Output1.0056.25
AC代码如下：
#include "iostream"#include "algorithm"#include "cstdio"#include "cmath"using namespace std;int main(int argc, char* argv[]){    double m,n,area;    double x1,x2,x3,x4,y1,y2,y3,y4,a[4],b[4];    while(cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4)    {        a[0]=x1;a[1]=x2;a[2]=x3;a[3]=x4;        b[0]=y1;b[1]=y2;b[2]=y3;b[3]=y4;        sort(a,a+4);        sort(b,b+4);        m=fabs(x1-x2)+fabs(x3-x4)-(a[3]-a[0]);        n=fabs(y1-y2)+fabs(y3-y4)-(b[3]-b[0]);        area=m*n;        if (m<0||n<0)        {            area=0;        }        printf("%.2lf/n",area);    }    return 0;}