P4251 [scoi2015] small convex play matrix (still a little confused)

https://www.luogu.com.cn/problem/P4251
Two points answer + Minimum match
You need to determine ： Can you find out at least n-k+1 Number , Make these numbers not greater than the current value and satisfy the condition .

seek n The number of k The minimum of a large number --------- Two points answer
Select a number per line ----------------------- Bipartite graph can be established
The first k The minimum of a large number ----------- Two points make a second k Large number , Find the maximum matching number , If the maximum match is greater than n-k, It means that this number can be even smaller , Reduce the boundary of the dichotomy .

#include <bits/stdc++.h>

using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=1e5+5;
int n,m,k,mp[300][300],link[maxn],head[maxn];
bool used[maxn];
int cnt,ma=0,mi=inf;
struct node
{
    
    int to,nxt;
}e[maxn];
void add(int from,int to)
{
    
    e[++cnt].to=to;
    e[cnt].nxt=head[from];
    head[from]=cnt;
}
bool dfs(int u)
{
    
    for(int i=head[u];i!=-1;i=e[i].nxt)
    {
    
        int v=e[i].to;
        if(!used[v])
        {
    
            used[v]=1;
            if(link[v]==-1||dfs(link[v]))
            {
    
                link[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
bool pd(int mid)
{
    
    memset(link,-1,sizeof(link));
    memset(head,-1,sizeof(head));
    for(int i=1;i<=n;i++)
    {
    
        for(int j=1;j<=m;j++)
        {
    
            if(mp[i][j]<=mid)
                add(i,j+n);
        }
    }
    int res=0;
    for(int i=1;i<=n;i++)
    {
    
        memset(used,0,sizeof(used));
        if(dfs(i))
            res++;
    }
    return res>=n-k+1;
}
int main()
{
    
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++)
    {
    
        for(int j=1;j<=m;j++)
        {
    
            scanf("%d",&mp[i][j]);
            ma=max(ma,mp[i][j]);
            mi=min(mi,mp[i][j]);
        }
    }
    int l=mi,r=ma,mid,ans;
    while(l<=r)
    {
    
        mid=(l+r)>>1;
        if(pd(mid))
            r=mid-1,ans=mid;
        else
            l=mid+1;
    }
    cout<<ans<<endl;
    return 0;
}