(Matching point) bipartite graph maximum matching

Portal : https://vjudge.net/contest/357926#problem/G

题目：On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (ai, bi), and the coordinates of the i-th blue point are (c_i, d_i).

A red point and a blue point can form a friendly pair when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point.

At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.

The meaning of problems: there are n points red and blue-n points in the plane, the red and blue dots between the pairs. If the points a and b, a red dot, a blue dot b, b in a lower left corner, i.e., x and y coordinates of a relatively small b, then these two points can be paired, the two paired and other points can not be paired, you can find the maximum number of pairing

Thinking: because when a, b, x and y are relatively small, these two points can be paired, i.e. between the two sides can be built Representative (herein labeled with flag bit array).

1  / * Maximum Matching: ① ② built FIG edged portion sequentially traverse the left point which determines matching ③find * / 
2  / * First Come, let let * /  
. 3 #include <the iostream>
 . 4  the using  namespace STD;
 . 5  const  int N = 110 ;
 . 6  int match [N]; // point corresponding to the right value 
. 7  BOOL VIS [N], in Flag [N] [N]; // VIS weight determination has not looked for 
. 8  struct Node {
 . 9      int X, Y;
 10 } R & lt [N], B [N]; // red and blue dots 
. 11  int n-;
 12 is  BOOL Find (int X) // to find blue red point corresponding to a point 
13 is  {
 14      for ( int I = . 1 ; I <= n-; I ++) // enumerate all points blue 
15      {
 16          IF (VIS [I! ] && in Flag [X] [I]) // the blue dots are not viewed and can be matched with the red dot 
. 17          {
 18 is              VIS [I] = . 1 ;
 . 19              IF (match [I] == 0 || Find ( match [I]))
 20 is              { // point not matching the blue through the blue dot || been matched by a red dot but may also match other 
21 is                  match [I] = X;
 22 is                  return  . 1;
23             }
24         }
25     }
26     return 0;
27 }
28 int main()//主函数
29 {
30     cin>>n;
31     for(int i=1;i<=n;i++) cin>>r[i].x>>r[i].y;
32     for(int i=1;i<=n;i++) cin>>b[i].x>>b[i].y;
33     for(int i=1;i<=n;i++)
34     {
35         for(int j=1;j<=n;j++)
36         {
37             if(r[i].x<b[j].x&&r[i].y<b[j].y)
38             {
39                 flag[i][j]=1;//两者可以配对 
40             }
41         }
42     }
43     long long ans=0;
44     for(int i=1;i<=n;i++)
45     {
46         memset(vis,0, The sizeof (VIS)); 
 47          IF (Find (I)) // maximum matching 
48          ANS ++ ;
 49       } 
 50      COUT ANS << << endl;
 51 is }

Code spicy chicken hh の