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Copy pathcc.cpp
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99 lines (89 loc) · 2.61 KB
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#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
template<class T> using uset = unordered_set<T>;
template<class T, class U> using umap = unordered_map<T, U>;
template<class T> using mset = multiset<T>;
template<class T, class U> using mmap = multimap<T, U>;
template<class T> using umset = unordered_multiset<T>;
template<class T, class U> using ummap = unordered_multimap<T, U>;
#define REP(i, n, m) for(ll i=n; i<(ll)m; ++i)
#define IREP(i, n, m) for(ll i=n-1; i>=m; --i)
#define rep(i, n) REP(i, 0, n)
#define irep(i, n) IREP(i, n, 0)
#define all(v) v.begin(), v.end()
#define vprint(v) for(auto e:v){cout<<e<<" ";};cout<<endl;
#define vvprint(vv) for(auto v:vv){vprint(v)};
class SegTree{
public:
ll size;
vll v;
vll lv;
SegTree(ll n){
this->size = n;
ll m = 1;
while(n>m) m *= 2;
this->v = vll(2*m-1, 0);
this->lv = vll(2*m-1, 0);
}
ll get(ll left, ll right, ll level=0, ll idx=0){
if(left>right) return 0;
ll step = (this->v.size()+1) >> (level+1);
ll lb = (idx-(1<<level)+1) * step;
ll ub = lb + step - 1;
if(lb==left && ub==right){
return v[idx];
}
else return max(get(left, min((lb+ub)/2, right), level+1, idx*2+1), get(max((lb+ub)/2+1, left), right, level+1, idx*2+2));
}
void update(ll x, ll e, ll level=0, ll idx=0){
ll step = (this->v.size()+1) >> (level+1);
ll lb = (idx-(1<<level)+1) * step;
ll ub = lb + step - 1;
if(lb==x && ub==x){
this->v[idx] = max(this->v[idx], e);
}
else{
this->v[idx] = max(this->v[idx], e);
ll mid = (lb+ub) / 2;
if(x<=mid) update(x, e, level+1, idx*2+1);
else update(x, e, level+1, idx*2+2);
}
}
};
int main(){
cin.tie(0);
ios::sync_with_stdio(false);
cout << setprecision(20);
ll N, K;
cin >> N >> K;
vll A(N), B(N);
rep(i, N) cin >> A[i] >> B[i];
vvll g(5000, vll(5000, 0));
rep(i, N) ++g[A[i]-1][B[i]-1];
K = min(5000ll, K+1);
vvll s(5000-K+1, vll(5000, 0));
rep(i, 5000){
ll t = 0;
rep(j, K) t += g[j][i];
s[0][i] = t;
REP(j, 1, 5000-K+1){
t += g[j+K-1][i] - g[j-1][i];
s[j][i] = t;
}
}
ll ans = 0;
rep(i, 5000-K+1){
ll t = 0;
rep(j, K) t += s[i][j];
ans = max(ans, t);
REP(j, 1, 5000-K+1){
t += s[i][j+K-1] - s[i][j-1];
ans = max(ans, t);
}
}
cout << ans << endl;
}