HDU 2056 Rectangles

RectanglesTime Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 23707    Accepted Submission(s): 7722Problem 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 Input1.00 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 12.50 Sample Output1.0056.25 Authorseeyou //参考了http://blog.sina.com.cn/s/blog_ab20767501017n1x.html#include<iostream>#include<iomanip>using namespace std;double max(double a,double b){    if(a>b)        return a;    else     return b;}double min(double a,double b){    if(a>b)        return b;    else      return a;}int main(){    double x1,x2,x3,x4,y1,y2,y3,y4,t;    while(cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4)    {        double area=0;        if(x1>x2) t=x1,x1=x2,x2=t;//保证了相对顺序        if(y1>y2) t=y1,y1=y2,y2=t;        if(x3>x4) t=x3,x3=x4,x4=t;        if(y3>y4) t=y3,y3=y4,y4=t;                 x1=max(x1,x3);//重合矩阵         y1=max(y1,y3);         x2=min(x2,x4);         y2=min(y2,y4);        if(x1>x2||y1>y2)            cout<<"0.00"<<endl;        else        {           area=(x2-x1)*(y2-y1);           cout<<setprecision(2)<<std::fixed<<area<<endl;        }    }    return 0;}