이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int ll
const int N = 2000 + 7;
const int K = 20;
int n;
int m;
int q;
vector<int> g[N];
pair<int, int> edges[N];
int dep[N];
int par[N];
int l[N];
int r[N];
int tab[K][N];
int tt = 0;
int sub[N];
pair<int, int> updates[N];
void build(int a, int p = 0) {
l[a] = ++tt;
tab[0][a] = p;
for (int k = 1; k < K; k++) {
tab[k][a] = tab[k - 1][tab[k - 1][a]];
}
sub[a] = 1;
for (auto &b : g[a]) {
if (b != p) {
par[b] = a;
dep[b] = dep[a] + 1;
build(b, a);
sub[a] += sub[b];
}
}
r[a] = tt;
}
int lca(int x, int y) {
if (dep[x] < dep[y]) {
swap(x, y);
}
for (int k = K - 1; k >= 0; k--) {
if (dep[x] - (1 << k) >= dep[y]) {
x = tab[k][x];
}
}
if (x == y) {
return x;
}
for (int k = K - 1; k >= 0; k--) {
if (tab[k][x] != tab[k][y]) {
x = tab[k][x];
y = tab[k][y];
}
}
return tab[0][x];
}
bool is_ancestor(int x, int y) { /// x ancestor of y
return l[x] <= l[y] && r[y] <= r[x];
}
struct PST{
struct Node {
int result;
Node* left;
Node* right;
Node (int val = 0) {
result = val;
left = right = nullptr;
}
void refresh() {
this->result = this->left->result + this->right->result;
}
};
Node* build(int from, int to) {
Node *node = new Node(0);
if (from < to) {
int mid = (from + to) / 2;
node->left = build(from, mid);
node->right = build(mid + 1, to);
}
return node;
}
Node* initialize(int n) {
return build(1, n);
}
Node* update(Node* base, int from, int to, int x, int val) {
if (from < to) {
int mid = (from + to) / 2;
Node* newNode = new Node();
(*newNode) = (*base);
if (x <= mid) {
newNode->left = update(base->left, from, mid, x, val);
} else {
newNode->right = update(base->right, mid + 1, to, x, val);
}
newNode->refresh();
return newNode;
} else {
return (new Node(base->result + val));
}
}
int query(Node* node, int from, int to, int x, int y) {
assert(from <= x && x <= y && y <= to);
if (from == x && to == y) {
return node->result;
} else {
int mid = (from + to) / 2;
if (x <= mid && y <= mid) {
return query(node->left, from, mid, x, y);
} else if (mid + 1 <= x && mid + 1 <= y) {
return query(node->right, mid + 1, to, x, y);
} else {
return query(node->left, from, mid,x , mid) + query(node->right, mid + 1, to, mid + 1, y);
}
}
}
Node* version[N];
int n;
void buildVersions(int n_) {
n = n_;
Node* part = initialize(n);
version[0] = part;
for(int i = 1; i <= m; i++) {
int val = updates[i].first;
int x = updates[i].second;
part = update(part, 1, n, l[x], +val);
if (r[x] + 1 <= n) {
part = update(part, 1, n, r[x] + 1, -val);
}
version[i] = part;
}
}
int sum(int when, int x) {
return query(version[when], 1, n, 1, l[x]);
}
int get_sum(int when, int x, int y) {
int z = lca(x, y);
return sum(when, x) + sum(when, y) - 2 * sum(when, z);
}
};
PST tree1, tree2;
signed main() {
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
/// freopen("input.txt", "r", stdin);
cin >> n >> m >> q;
for (int i = 1; i < n; i++) {
int x, y;
cin >> x >> y;
g[x].push_back(y);
g[y].push_back(x);
edges[i] = {x, y};
}
build(1);
for (int i = 1; i <= m; i++) {
ll road, val;
cin >> road >> val;
int x = edges[road].first;
int y = edges[road].second;
if (dep[x] < dep[y]) {
swap(x, y);
}
updates[i] = {val, x};
}
sort(updates + 1, updates + m + 1);
tree1.buildVersions(n);
for (int i = 1; i <= m; i++) {
updates[i].first = 1;
}
tree2.buildVersions(n);
while (q--) {
ll s, t, x, y;
cin >> s >> t >> x >> y;
int low = 1, high = m, sol = 0;
while (low <= high) {
int mid = (low + high) / 2;
if (tree1.get_sum(mid, s, t) <= y) {
sol = mid;
low = mid + 1;
} else {
high = mid - 1;
}
}
int z = lca(s, t);
int payed = tree2.get_sum(sol, s, t);
int total = tree2.get_sum(m, s, t);
int keep = x - (total - payed);
if (keep < 0) {
keep = -1;
}
cout << keep << "\n";
}
return 0;
}
컴파일 시 표준 에러 (stderr) 메시지
currencies.cpp: In function 'int main()':
currencies.cpp:206:9: warning: unused variable 'z' [-Wunused-variable]
206 | int z = lca(s, t);
| ^
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