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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#537627 | #7523. Partially Free Meal | OIer_kzc# | TL | 1ms | 7880kb | C++17 | 2.8kb | 2024-08-30 16:58:45 | 2024-08-30 16:58:45 |
Judging History
answer
#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
#include <numeric>
#define LOG(FMT...) fprintf(stderr, FMT)
#define eb emplace_back
using namespace std;
typedef long long LL;
constexpr int N = 200005, TC = 30 * N;
int n;
struct Pair {
int a, b, k;
} w[N];
bool cmpA(const Pair &x, const Pair &y) {
return x.a < y.a;
}
bool cmpB(const Pair &x, const Pair &y) {
return x.b < y.b;
}
#define L (tr[z].l)
#define R (tr[z].r)
struct Tree {
int l, r, c; LL v;
} tr[TC];
int rt[N], ids;
void mdf(int &z, int x, int c, int l = 0, int r = n) {
z = ++ids;
tr[z].v = c;
tr[z].c = 1;
if (l == r) {
return;
}
int mid = l + r >> 1;
if (x <= mid) {
mdf(L, x, c, l, mid);
} else {
mdf(R, x, c, mid + 1, r);
}
}
void merge(int &y, int x, int l = 0, int r = n) {
if (!x) {
return;
}
if (!y) {
y = x;
return;
}
tr[y].v += tr[x].v;
tr[y].c += tr[x].c;
if (l == r) {
return;
}
int mid = l + r >> 1;
merge(tr[y].l, tr[x].l, l, mid);
merge(tr[y].r, tr[x].r, mid + 1, r);
}
struct Dat {
int c; LL d;
bool operator < (const Dat &t) const {
return c < t.c || c == t.c && d < t.d;
}
Dat operator + (int tc) const {
return (Dat){c + tc, d};
}
} q[N];
Dat qry(int z, LL v, int l = 0, int r = n) {
if (l == r) {
return (Dat){0, v};
}
int mid = l + r >> 1;
if (tr[L].v <= v) {
return qry(R, v - tr[L].v, mid + 1, r) + tr[L].c;
}
return qry(L, v, l, mid);
}
LL res[N];
void solve(int l, int r, LL vl, LL vr, const vector<int> &v) {
if (l > r) {
return;
}
if (vl == vr) {
for (int k = l; k <= r; ++k) {
res[k] = vr;
}
return;
}
LL md = (vl + vr) >> 1;
int maxc = 0;
for (int i = 0; i < v.size(); ++i) {
int x = v[i];
q[i] = qry(rt[x], md - w[x].b);
maxc = max(maxc, q[i].c);
}
vector<int> nv;
Dat t = (Dat){-1, -1};
for (int i = 0; i < v.size(); ++i) {
if (!(q[i] < t)) {
nv.eb(v[i]);
t = q[i];
}
}
solve(l, maxc, vl, md, v);
nv.clear();
t = (Dat){-1, -1};
for (int i = v.size() - 1; ~i; --i) {
if (t < q[i]) {
nv.eb(v[i]);
t = q[i];
}
}
reverse(nv.begin(), nv.end());
solve(maxc + 1, r, md + 1, vr, nv);
}
int main() {
scanf("%d", &n);
LL suma = 0; int maxb = 0;
for (int i = 1; i <= n; ++i) {
scanf("%d%d", &w[i].a, &w[i].b);
suma += w[i].a;
maxb = max(maxb, w[i].b);
}
sort(w + 1, w + n + 1, cmpA);
for (int i = 1; i <= n; ++i) {
w[i].k = i - 1;
}
sort(w + 1, w + n + 1, cmpB);
for (int i = 1; i <= n; ++i) {
mdf(rt[i], w[i].k, w[i].a);
merge(rt[i], rt[i - 1]);
}
vector<int> v(n);
iota(v.begin(), v.end(), 1);
solve(1, n, 0, suma + maxb, v);
for (int i = 1; i <= n; ++i) {
printf("%lld\n", res[i]);
}
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 1ms
memory: 7880kb
input:
3 2 5 4 3 3 7
output:
7 11 16
result:
ok 3 lines
Test #2:
score: -100
Time Limit Exceeded
input:
200000 466436993 804989151 660995237 756645598 432103296 703610564 6889895 53276988 873617076 822481192 532911431 126844295 623111499 456772252 937464699 762157133 708503076 786039753 78556972 5436013 582960979 398984169 786333369 325119902 930705057 615928139 924915828 506145001 164984329 208212435...