AtCoder Regular Contest 092 C

Problem Statement

On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i,b_i), 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.

Constraints

• All input values are integers.
• 1≤N≤100
• 0≤a_i,b_i,c_i,d_i<2N
• a₁,a₂,…,a_N,c₁,c₂,…,c_N are all different.
• b₁,b₂,…,b_N,d₁,d₂,…,d_N are all different.

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Input

Input is given from Standard Input in the following format:

N
a1 b1
a2 b2
:
aN bN
c1 d1
c2 d2
:
cN dN

Output

Print the maximum number of friendly pairs.

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Sample Input 1

Copy

3
2 0
3 1
1 3
4 2
0 4
5 5

Sample Output 1

Copy

2

For example, you can pair (2,0) and (4,2), then (3,1) and (5,5).

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Sample Input 2

Copy

3
0 0
1 1
5 2
2 3
3 4
4 5

Sample Output 2

Copy

2

For example, you can pair (0,0) and (2,3), then (1,1) and (3,4).

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Sample Input 3

Copy

2
2 2
3 3
0 0
1 1

Sample Output 3

Copy

0

It is possible that no pair can be formed.

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Sample Input 4

Copy

5
0 0
7 3
2 2
4 8
1 6
8 5
6 9
5 4
9 1
3 7

Sample Output 4

Copy

5

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Sample Input 5

Copy

5
0 0
1 1
5 5
6 6
7 7
2 2
3 3
4 4
8 8
9 9

Sample Output 5

Copy

4

题意

给你n个二维坐标点A，再给n个二维坐标点B，如果B的x和y均大于A，这两个点可以匹配，求A的最大匹配

题解

一道二分图匹配的题，算是比较基础的题

当然这个题可以用贪心从小的开始，每次都选择B里与其最相近的元素

代码

#include<bits/stdc++.h>
using namespace std;

const int N=310;
pair<int,int> p[N];
int vis[N],match[N];
int n;
int Find(int u)
{
    for(int i=n+1;i<=n+n;i++)
    {
        if(p[i].first>p[u].first&&p[i].second>p[u].second&&!vis[i])
        {
            vis[i]=1;
            if(!match[i]||Find(match[i]))
            {
                match[i]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    cin>>n;
    for(int i=1;i<=n+n;i++)
        cin>>p[i].first>>p[i].second;
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        memset(vis,0,sizeof(vis));
        if(Find(i))
            ans++;
    }
    cout<<ans;
    return 0;
}