#include <cstdio>
#include <cstring>
#include <cassert>
#include <string>
#include <deque>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <iostream>
#include <utility>
using namespace std;
using ll=long long;
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using iset = tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update>;
#define N 200005
#define ALL(x) x.begin(), x.end()
int n, q, C, ans[N];
iset t[N<<1];
void build(int v, int l, int r)
{
int m = (l+r)/2, vl = v+1, vr = v+(m-l+1)*2;
if (l == r)
;
else
{
build(vl, l, m), build(vr, m+1, r);
for (auto x : t[vl]) t[v].insert(x);
}
}
void Ins(int v, int l, int r, int p, int k)
{
if (r < p || p < l) return;
if (l != r)
{
int m = (l+r)/2, vl = v+1, vr = v+(m-l+1)*2;
Ins(vl, l, m, p, k), Ins(vr, m+1, r, p, k);
}
t[v].insert(make_pair(k, ++C));
}
int Qry(int v, int l, int r, int x, int y, int k)
{
if (r < x || y < l) return 0;
if (x <= l && r <= y) return t[v].size() - t[v].order_of_key(*t[v].lower_bound(make_pair(k, 0)));
int m = (l+r)/2, vl = v+1, vr = v+(m-l+1)*2;
return Qry(vl, l, m, x, y, k) + Qry(vr, m+1, r, x, y, k);
}
struct qry
{
int x, y, z, i, cx;
bool operator<(const qry &a) const { return z > a.z; }
} b[N];
struct ppl
{
int x, y, cx;
bool operator<(const ppl &a) const { return x+y > a.x + a.y; }
} a[N];
void compress()
{
vector<int> v;
for (int i = 0; i < n; ++i) v.push_back(a[i].x);
for (int i = 0; i < q; ++i) v.push_back(b[i].x);
sort(ALL(v));
for (int i = 0; i < n; ++i) a[i].cx = lower_bound(ALL(v), a[i].x) - v.begin() + 1;
for (int i = 0; i < q; ++i) b[i].cx = lower_bound(ALL(v), b[i].x) - v.begin() + 1;
}
int main()
{
cin.tie(0)->sync_with_stdio(0);
cin >> n >> q;
for (int i = 0; i < n; ++i) cin >> a[i].x >> a[i].y;
for (int i = 0; i < q; ++i) cin >> b[i].x >> b[i].y >> b[i].z, b[i].i = i;
compress();
sort(a, a+n);
sort(b, b+q);
int j = 0;
for (int i = 0; i < q; ++i)
{
for (; j < n && a[j].x + a[j].y >= b[i].z; ++j)
Ins(0, 1, N, a[j].cx, a[j].y);
ans[b[i].i] = Qry(0, 1, N, b[i].cx, N-1, b[i].y);
}
for (int i = 0; i < q; ++i) printf("%d\n", ans[i]);
return 0;
}
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
21 ms |
42076 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
1744 ms |
162332 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
1744 ms |
162332 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
결과 |
실행 시간 |
메모리 |
Grader output |
1 |
Incorrect |
21 ms |
42076 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |