Rectangles

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 1 #include<iostream>
 2 #include<iomanip>
 3 using namespace std;
 4 
 5 class p
 6 {
 7 public:
 8     double x;
 9     double y;
10     friend istream& operator >>(istream& in, p& x);
11 };
12 
13 istream& operator >>(istream& in, p& a)
14 {
15     in >> a.x >> a.y;
16     return in;
17 }
18 void swap(p& a, p& b)
19 {
20     if(a.x > b.x)
21     {
22        double t = a.x;
23        a.x = b.x;
24        b.x = t;
25     }
26     if(a.y > b.y)
27     {
28         double t = a.y;
29         a.y = b.y;
30         b.y = t;
31     }
32 }
33 int main()
34 {
35     p a, b, c, d;
36     while(cin >> a >> b >> c >> d)
37     {
38         swap(a, b);
39         swap(c, d);
40         double area = 0.00;
41         p m, n;
42         m.x = a.x > c.x ? a.x : c.x;
43         m.y = a.y > c.y ? a.y : c.y;
44         n.x = b.x > d.x ? d.x : b.x;
45         n.y = b.y > d.y ? d.y : b.y;
46         cout.setf(ios::fixed);
47         cout.precision(2);
48         area = (m.x > n.x || m.y > n.y) ? 0 : (n.x-m.x)*(n.y-m.y);
49         cout << area << endl;
50     }
51     return 0;
52 }

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

View Code

 1 #include<iostream>
 2 #include<cmath>
 3 #include<algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     double x1,y1,x2,y2,x3,y3,x4,y4;
 8     double x[4],y[4];
 9     double s,w,h;
10     while(cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4)
11     {
12         x[0]=x1;
13         x[1]=x2;
14         x[2]=x3;
15         x[3]=x4;
16         y[0]=y1;
17         y[1]=y2;
18         y[2]=y3;
19         y[3]=y4;
20         sort(x,x+4);
21         sort(y,y+4);
22         w=fabs(x2-x1)+fabs(x4-x3)-(x[3]-x[0]);
23         h=fabs(y2-y1)+fabs(y4-y3)-(y[3]-y[0]);
24         s=w*h;
25         if(w<=0 || h<=0)s=0.00;
26         cout.setf(ios::fixed);
27         cout.precision(2);
28         cout << s << endl;
29     }
30     return 0;
31 }