Problem: Example

Input

5 5 4
6
3 2
4 2
5 2
1 4
3 4
5 4


Output

6

Solution (in CPP):

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

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcount

#define INF INT_MAX/3
#define mod 1000000007

int w,h,k,n,m;
vector<int> ps[10010];
int dp[5005];

int main(){
  cin>>w>>h>>k;
  cin>>n;
  rep(i,n){
    int a,b;
    cin>>a>>b;
    a--;b--;
    ps[b].push_back(a);
  }
  rep(i,h)sort(all(ps[i]));
  m=(w-1)/2+1;
  if(k<m){
    cout<<-1<<endl;
    return 0;
  }
  k-=m;
  int res=0;
  int cnt=0;
  for(int i=1;i<h;i+=2){
    rep(j,m+1)dp[j]=INF;
    dp[0]=0;
    rep(j,m){
      int idx1=lower_bound(all(ps[i]),j*2)-ps[i].begin();
      int idx2=lower_bound(all(ps[i]),j*2+1)-ps[i].begin();
      int v1=(idx1<ps[i].size()&&ps[i][idx1]==(j*2)?1:0);
      int v2=(idx2<ps[i].size()&&ps[i][idx2]==(j*2+1)?1:0);
      minch(dp[j+1],dp[j]+v1);
      if(j+2<=m)minch(dp[j+2],dp[j]+v2*2);
    }
    cnt+=dp[m];
    res+=dp[m]*2+(m-dp[m])*1;
  }
  if(cnt>k){
    res-=2*k;
  }else{
    res-=2*cnt+(k-cnt);
  }
  cout<<max(0,res)<<endl;
  return 0;
}