結果

問題 No.2311 [Cherry 5th Tune] Cherry Month
ユーザー square1001square1001
提出日時 2023-03-24 23:20:48
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,288 bytes
コンパイル時間 1,585 ms
コンパイル使用メモリ 94,336 KB
実行使用メモリ 817,872 KB
最終ジャッジ日時 2024-12-15 18:59:03
合計ジャッジ時間 22,645 ms
ジャッジサーバーID
(参考情報)
judge3 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
other AC * 23 WA * 25 MLE * 3
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ソースコード

diff #
プレゼンテーションモードにする

#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
class query {
public:
int tp, t, x, y;
query() : tp(-1), t(-1), x(-1), y(-1) {}
query(int tp_, int t_, int x_, int y_) : tp(tp_), t(t_), x(x_), y(y_) {}
bool operator<(const query& q) const {
return t < q.t;
}
};
int main() {
// step #1. input
cin.tie(0);
ios::sync_with_stdio(false);
int N;
cin >> N;
vector<long long> A(N);
for (int i = 0; i < N; i++) {
cin >> A[i];
}
int M;
cin >> M;
vector<query> qs(M);
for (int i = 0; i < M; i++) {
int tp, x, y;
cin >> tp >> x >> y;
qs[i] = query(tp, i * 2 + 1, x - 1, tp == 1 ? y - 1 : y);
}
int Q;
cin >> Q;
qs.resize(M + Q);
for (int i = 0; i < Q; i++) {
int t, x;
cin >> t >> x;
qs[M + i] = query(5, t * 2, x - 1, i);
}
sort(qs.begin(), qs.end());
// step #2. renumber vertices
vector<vector<int> > group(N);
vector<int> root(N), id(N), idl(N), idr(N);
for (int i = 0; i < N; i++) {
group[i] = { i };
root[i] = i;
id[i] = i;
}
for (query q : qs) {
if (q.tp == 1) {
int rx = root[q.x];
int ry = root[q.y];
if (rx != ry) {
if (group[rx].size() < group[ry].size()) {
swap(rx, ry);
}
group[rx].insert(group[rx].end(), group[ry].begin(), group[ry].end());
int delta = idr[rx] - idl[rx];
for (int j : group[ry]) {
id[j] += delta;
}
idl[ry] += delta;
idr[ry] += delta;
for (int j : group[ry]) {
root[j] = rx;
}
}
}
}
for (query &q : qs) {
if (q.tp == 1) {
q.x = id[q.x];
q.y = id[q.y];
}
else {
q.x = id[q.x];
}
}
vector<long long> TA = A;
for (int i = 0; i < N; i++) {
A[id[i]] = TA[i];
}
// step #3. find left/right of update queries
for (int i = 0; i < N; i++) {
group[i] = { i };
root[i] = i;
idl[i] = i;
idr[i] = i + 1;
}
vector<int> l(M + Q, -1), r(M + Q, -1);
for (int i = 0; i < M + Q; i++) {
query q = qs[i];
if (q.tp == 1) {
int rx = root[q.x];
int ry = root[q.y];
if (rx != ry) {
if (group[rx].size() < group[ry].size()) {
swap(rx, ry);
}
group[rx].insert(group[rx].end(), group[ry].begin(), group[ry].end());
int delta = idr[rx] - idl[rx];
idl[rx] = min(idl[rx], idl[ry]);
idr[rx] = max(idr[rx], idr[ry]);
for (int j : group[ry]) {
root[j] = rx;
}
}
}
if (q.tp == 4) {
l[i] = idl[root[q.x]];
r[i] = idr[root[q.x]];
}
}
// step #4. find all high-degree vertices
vector<int> deg(N, 0);
for (query q : qs) {
if (q.tp == 1) {
deg[q.x] += 1;
deg[q.y] += 1;
}
}
int Z = 0;
vector<int> vert_id(N, -1);
for (int i = 0; i < N; i++) {
if (1LL * deg[i] * deg[i] >= N) {
vert_id[i] = Z;
Z += 1;
}
}
vector<vector<int> > flag(N, vector<int>(Z, -1));
// step #5. binary indexed tree
vector<long long> bit(N + 1);
auto add = [&](int pos, int val) {
for (int i = pos + 1; i <= N; i += i & (-i)) {
bit[i] += val;
}
};
auto getsum = [&](int pos) {
long long answer = 0;
for (int i = pos; i >= 1; i -= i & (-i)) {
answer += bit[i];
}
return answer;
};
// step #6. process queries
vector<vector<int> > g(N);
vector<long long> val(N, 0), answer(Q, -1);
vector<vector<long long> > bigval(M + Q + 1);
bigval[0] = vector<long long>(Z, 0);
for (int i = 0; i < M + Q; i++) {
bigval[i + 1] = bigval[i];
query q = qs[i];
if (q.tp == 1) {
if (vert_id[q.x] == -1) {
g[q.x].push_back(q.y);
}
else if (flag[q.y][vert_id[q.x]] == -1) {
flag[q.y][vert_id[q.x]] = i;
}
if (vert_id[q.y] == -1) {
g[q.y].push_back(q.x);
}
else if (flag[q.x][vert_id[q.y]] == -1) {
flag[q.x][vert_id[q.y]] = i;
}
}
if (q.tp == 2) {
val[q.x] += q.y;
}
if (q.tp == 3) {
if (vert_id[q.x] == -1) {
for (int j : g[q.x]) {
val[j] += q.y;
}
}
else {
bigval[i + 1][vert_id[q.x]] += q.y;
}
val[q.x] += q.y;
}
if (q.tp == 4) {
add(l[i], q.y);
add(r[i], -q.y);
}
if (q.tp == 5) {
long long subval = val[q.x];
for (int j = 0; j < Z; j++) {
if (flag[q.x][j] != -1) {
subval += bigval[i + 1][j] - bigval[flag[q.x][j]][j];
}
}
subval += getsum(q.x + 1);
answer[q.y] = max(A[q.x] - subval, 0LL);
}
}
// step #7. output
for (int i = 0; i < Q; i++) {
cout << answer[i] << '\n';
}
return 0;
}
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