观察特点

杭电2056：

Rectangles

博客：Kings

#include<iostream>
#include<iomanip>
using namespace std;

class p
{
public:
    double x;
    double y;
    friend istream& operator >>(istream& in, p& x);
};

istream& operator >>(istream& in, p& a)
{
    in >> a.x >> a.y;
    return in;
}
void swap(p& a, p& b)
{
    if(a.x > b.x)
    {
       double t = a.x;
       a.x = b.x;
       b.x = t;
    }
    if(a.y > b.y)
    {
        double t = a.y;
        a.y = b.y;
        b.y = t;
    }
}
int main()
{
    p a, b, c, d;
    while(cin >> a >> b >> c >> d)
    {
        swap(a, b);
        swap(c, d);
        double area = 0.00;
        p m, n;
        m.x = a.x > c.x ? a.x : c.x;
        m.y = a.y > c.y ? a.y : c.y;
        n.x = b.x > d.x ? d.x : b.x;
        n.y = b.y > d.y ? d.y : b.y;
        cout.setf(ios::fixed);
        cout.precision(2);
        area = (m.x > n.x || m.y > n.y) ? 0 : (n.x-m.x)*(n.y-m.y);
        cout << area << endl;
    }
    return 0;
}

作图分析可知若两矩形相交相交两交点坐标应该是上述代码中的（x1,y1） (x2,y2)，并且满足x2>x1 y2>y1，反之则两矩形不想交面积为0

#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,w,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);
        w=fabs(x2-x1)+fabs(x4-x3)-(x[3]-x[0]);
        h=fabs(y2-y1)+fabs(y4-y3)-(y[3]-y[0]);
        s=w*h;
        if(w<=0 || h<=0)s=0.00;
        cout.setf(ios::fixed);
        cout.precision(2);
        cout << s << endl;
    }
    return 0;
}