QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#321620 | #7563. Fun on Tree | bobbilyking | ML | 0ms | 20004kb | C++17 | 14.6kb | 2024-02-05 01:14:47 | 2024-02-05 01:14:48 |
Judging History
answer
// this is the battefield to ac on probelm F
#pragma GCC target ("avx2")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#include<bits/stdc++.h>
#include<math.h>
using namespace std;
typedef long long int ll;
typedef long double ld;
typedef pair<ll, ll> pl;
typedef vector<ll> vl;
#define FD(i, r, l) for(ll i = r; i > (l); --i)
#define K first
#define V second
#define G(x) ll x; cin >> x;
#define GD(x) ld x; cin >> x;
#define GS(s) string s; cin >> s;
#define EX(x) { cout << x << '\n'; exit(0); }
#define A(a) (a).begin(), (a).end()
#define F(i, l, r) for (ll i = l; i < (r); ++i)
const ll N = 2e5 + 10;
ll tin[N], tout[N];
vector<pl> adj[N];
void dfs(ll i, ll p) {
static int time = 0;
tin[i] = time++;
for (auto [x, _]: adj[i]) if (x-p) dfs(x, i);
tout[i] = time;
}
namespace lca {
const ll L = 20;
ll dep[N], par[N][L], distFromRoot[N];
void lcadfs(ll i, ll p, ll pw) {
static int time = 0;
dep[i] = dep[p] + 1;
par[i][0] = p;
distFromRoot[i] = pw;
tin[i] = time++;
F(l, 1, L) {
par[i][l] = par[par[i][l - 1]][l - 1];
}
for(auto [j, w]: adj[i]) if(j - p) lcadfs(j, i, pw + w);
tout[i] = time;
}
ll lca(ll a, ll b) {
if(dep[a] < dep[b]) swap(a, b);
FD(l, L - 1, -1) if((dep[a] - dep[b]) >> l) a = par[a][l];
if(a == b) return a;
FD(l, L - 1, -1) if(par[a][l] - par[b][l]) a = par[a][l], b = par[b][l];
return par[a][0];
}
ll dist(ll a, ll b){
return distFromRoot[a] + distFromRoot[b] - 2 * distFromRoot[lca(a, b)];
}
}
typedef pl T;
typedef ll U;
// combining segtree nodes a and b
T f(T a, T b) {
return max(a, b);
}
// applying updates a and b (in that order)
U g(U b, U a) { return a + b; }
// applying update b to segtree node a
T h(U b, T a) {return {a.K + b, a.V};}
T idT = {-1e18, -1};
struct lztree {
ll n, nL;
vector<T> t;
U idU = 0;
vector<U> d;
void calc(ll p) { t[p] = h(d[p], f(t[p * 2], t[p * 2 + 1])); }
void apply(ll p, U v) {
t[p] = h(v, t[p]);
if(p < n) d[p] = g(v, d[p]);
}
void push(ll p) {
p += n;
FD(s, nL, 0) {
ll i = p >> s;
if(d[i] != idU) {
apply(i * 2, d[i]);
apply(i * 2 + 1, d[i]);
d[i] = idU;
}
}
}
#define rein(p) lower_bound(A(mytins), p) - mytins.begin()
// difference between tshi: i reindexx before doing shit
// void modify(ll p, T v) {
// p = rein(p);
// push(p);
// t[p += n] = v;
// while(p > 1) calc(p /= 2);
// }
void modify(ll l, ll r, U v) {
if (v == idU) return;
l = rein(l), r = rein(r);
if (l >= r) return;
push(l), push(r - 1);
bool cl = false, cr = false;
for(l += n, r += n; l < r; l /= 2, r /= 2) {
if(cl) calc(l - 1);
if(cr) calc(r);
if(l & 1) apply(l++, v), cl = true;
if(r & 1) apply(--r, v), cr = true;
}
for(--l; r; l /= 2, r /= 2) {
if(cl) calc(l);
if(cr) calc(r);
}
}
T query() const { return t[1]; }
lztree() {}
// This lazy tree wants to effectively query tin/tout ranges.
// So reindex everything by tin; it is the responsibility of the passer to pass in correct tin/tout ranges.
// (point modifies pass in tin ranges)
// pass in pair of (node, value)
vl mytins;
lztree(vector<pl> pts) {
assert(pts.size());
sort(A(pts), [&](const pl &a, const pl& b) {
return tin[a.K] < tin[b.K];
});
n = pts.size();
nL = 32 - __builtin_clz(n);
mytins.resize(n);
t.resize(2*n, idT);
d.resize(n, idU);
F(i, 0, pts.size()) {
mytins[i] = tin[pts[i].K];
t[i+n] = {pts[i].V, -pts[i].K};
}
FD(i, n-1, -1) t[i] = f(t[i*2], t[i*2+1]);
}
};
ll a[N];
ll sz[N];
struct centroid_tree {
// operates on the assumption that both c[0] and c[1] are filled, or neither are filled
// (binary tree)
ll n;
ll centroid;
centroid_tree* c[2] = {nullptr, nullptr};
pl times = {1e9, -1e9};
ll lazy = 0;
// inefficient way of building for now
// this is so ugly
lztree seg;
centroid_tree(ll single) : n(1) {
centroid = single;
tie(times.K, times.V) = make_tuple(tin[single], tout[single]);
// This is VERY important since this
// dumb segtree cannot even support point update queries,
// only subtree...
// Point update queries are hard since under edge decomposition,
// many, many nodes can share the same centroid.
seg = lztree({pair(single, -a[single])});
}
centroid_tree(vector<pair<pl, ll>> edges,
centroid_tree *_par = nullptr) : n(edges.size() + 1) {
for (auto [a, w]: edges) {
auto [x, y] = a;
adj[x].clear(), adj[y].clear();
times.K = min(times.K, tin[x]);
times.V = max(times.V, tout[x]);
times.K = min(times.K, tin[y]);
times.V = max(times.V, tout[y]);
}
for (auto [a, w]: edges){
auto [x, y] = a;
adj[x].emplace_back(y, w), adj[y].emplace_back(x, w);
}
auto do_distances = [&](){
vector<pl> distances;
auto dfs = [&](auto &&self, ll i, ll p, ll d) -> void {
distances.emplace_back(i, d - a[i]);
for (auto [x, w]: adj[i]) if (x-p) self(self, x, i, d + w);
// our implementation fails with repeating edges... ffs...
};
// Now this is a hack if i've ever seen one.
// Again, we care about the distance from us to parent's centroid, not to our current centroid,
// because we want to do the flipping thing.
// For the root node, well uh, just do the normal centroid ig, everything should work out under evaluate?
// not sure abt this, revisit TODO
ll pcent = _par ? _par->centroid : centroid;
dfs(dfs, pcent, -1, 0);
seg = lztree(move(distances));
};
/*
do some processing with the adjlist here
*/
if (n == 3) {
ll x;
// casework on which one appears twice
if (edges[0].K.K == edges[1].K.K || edges[0].K.K == edges[1].K.V) x = edges[0].K.K;
else x = edges[0].K.V;
auto &v = adj[x];
vector<pair<pl, ll>> nedges[2];
F(i, 0, 2) nedges[i].emplace_back(pair(pair(v[i].K, x), v[i].V));
centroid = x;
F(i, 0, 2) c[i] = new centroid_tree(v[i].K);
do_distances();
return;
} else if (n == 2) {
c[0] = new centroid_tree(edges[0].K.K);
c[1] = new centroid_tree(edges[0].K.V);
centroid = c[0]->centroid; // this can just be whatever, doesn't matter
do_distances();
return;
}
assert(n != 1);
function<ll(ll, ll)> getSize = [&](ll i, ll p) {
sz[i] = 1;
for (auto [x, _]: adj[i]) if (x-p)
sz[i] += getSize(x, i);
return sz[i];
};
function<ll(ll, ll)> findcent = [&](ll i, ll p) {
for (auto [x, _]: adj[i]) if (x-p and sz[x] > n/2)
return findcent(x, i);
return i;
};
getSize(edges[0].K.K, -1);
centroid = findcent(edges[0].K.K, -1);
do_distances();
getSize(centroid, -1);
ll tot = 0;
vector<pair<pl, ll>> nedge[2];
for (auto [x, _]: adj[centroid]) {
ll idx;
if (tot + sz[x] <= 2 * n / 3) {
tot += sz[x];
idx = 0;
} else idx = 1;
function<void(ll, ll)> add_edges = [&](ll i, ll p) {
for (auto [x, w]: adj[i]) if (x-p) add_edges(x, i);
else {
nedge[idx].emplace_back(pair(i, p), w);
}
};
add_edges(x, centroid);
}
F(i, 0, 2) {
assert(nedge[i].size());
c[i] = new centroid_tree(move(nedge[i]), this);
}
}
bool leaf() const {
return c[0] == nullptr;
}
void push() {
if (!lazy) return;
// full application
seg.modify(tin[1], tout[1], lazy);
F(i, 0, 2) if (c[i]) c[i]->lazy += lazy;
lazy = 0; // reset to identity
}
// TODO: FILL OUT WITH LAZY SEGTREE OR SOMETH - DONE
pl evaluate() {
push();
return seg.query();
}
// segtree feels kinda wasteful here? all we are doing is range updating and then range querying aggregates over the whole
// tree. perhaps there is some kind of better data structure for this.
void update(ll node_index, ll value) {
// cout << "WTF " << centroid << " " << node_index << " " << value << " : " << n << endl;
push();
// THIS is necessarily true. THIS is a sufficient condition, 100%. Since this PRECISELY corresponds to
// "Everything in this centroid subtree are the nodes that will precisely get affected by me".
// Can even be a subset of my subtree, this will still hold = works.
// cout << tin[node_index] << " " << tout[node_index] << " | " << times.K << " " << times.V << endl;
// for (auto x: care) cout << x << " "; cout << endl;
if (tin[node_index] <= times.K and times.V <= tout[node_index]) {
lazy += value;
return;
}
// This only works for a centroid tree that has no ancestors of the descired node_index
if (tout[node_index] <= times.K || times.V <= tin[node_index]) return;
// If I have ancestors of my node_index, this will fail horribly.
// But if I have a node as my ancestor, I need to check if node_index is actually visited by this set of
// centroid nodes.
// If if its, then it's partial; go recurse.
// OTHERWISE, I must not affect anything in the subtree of node_index, therefore we are good to return here.
auto pos = lower_bound(A(seg.mytins), tin[node_index]);
bool contains = pos != seg.mytins.end() and *pos == tin[node_index];
if (contains) {
// go do ur recursion bro
} else return; // stop here
// cout << "ACTUALLY APPLYING LOL! " << endl;
// for (auto x: care) cout << x << " "; cout << endl;
// TODO - DONE
seg.modify(tin[node_index], tout[node_index], value);
// segment tree::update(node_index, node_value) or something
F(i, 0, 2) {
assert(c[i]);
c[i]->update(node_index, value);
}
}
pl query(ll node_index) {
push();
// cout << "is my universe actually correct...? " << endl;
// cout << "QUERYING " << centroid << endl;
// for (auto x: care) cout << x << " "; cout << endl;
// for (auto x: care) cout << seg.query(tin[x], tin[x]+1) << " "; cout << endl;
// cout << "wat? " << evaluate() << endl;
// cout << "======================" << endl;
if (node_index == centroid) {
T res = idT;
if (!leaf()) {
F(i, 0, 2) res = f(res, h((c[i]->leaf() ? lca::dist(c[i]->centroid, centroid) : 0), c[i]->evaluate()));
} else {
res = evaluate();
}
// warning: above edge case is kinda hacky, but ig the way i set it up makes sense like that
// This works since both children will point to me, the centroid.
// cout << "RETURNING ZERO " << res << endl;
return res;
}
ll directContribution = lca::dist(node_index, centroid);
// ll directContribution = f(node_index, centroid); path length, etc.
T ans = idT;
// cout << " YA " << node_index << " " << centroid << " ; " << n << endl;
// cout << "? " << directContribution << endl;
ll tried = 0;
F(i, 0, 2) {
assert(c[i]);
auto [tl, tr] = c[i]->times;
auto pos = lower_bound(A(c[i]->seg.mytins), tin[node_index]);
bool contains = pos != c[i]->seg.mytins.end() and *pos == tin[node_index];
if (!contains) {
ans = f(ans, h(directContribution, c[i]->evaluate()));
// cout << "THE OTHER UNIVERSE???? " << endl;
// for (auto x: c[i]->care) cout << x << " "; cout << endl;
// for (auto x: c[i]->care) cout << c[i]->seg.query(tin[x], tin[x]+1) << " "; cout << endl;
// cout << "WTF " << c[i]->evaluate() << " " << directContribution << endl;
} else {
// go further down the tree; assert that the condition is true for debugging purposes
assert(contains);
ans = max(ans, c[i]->query(node_index));
tried++;
}
}
assert(tried == 1);
// cout << "ANSWER: " << ans << endl;
return ans;
}
~centroid_tree() {
F(i, 0, 2) if (c[i]) delete c[i];
}
};
int main(){
// freopen("a.in", "r", stdin);
// freopen("a.out", "w", stdout);
ios_base::sync_with_stdio(false); cin.tie(0);
cout << fixed << setprecision(20);
G(n) G(q)
F(i ,1, n+1) cin >> a[i];
vector<pair<pl, ll>> tadj;
F(i, 2, n+1) {
G(p) G(w)
adj[p].emplace_back(i, w);
tadj.emplace_back(pair(i, p), w);
}
dfs(1, 1);
lca::lcadfs(1, 1, 0);
centroid_tree c(tadj);
ll updateCount = 0;
while(q--) {
G(x) G(y) G(v)
// ll crit = 4;
// if (++updateCount == crit) cout << "****************************************************************************" << endl;
c.update(y, -v);
auto [a, b] = c.query(x);
cout << -b << " " << a << endl;
// if (updateCount == crit) cout << "****************************************************************************" << endl;
}
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 20004kb
input:
6 6 1 1 4 5 1 4 1 5 2 0 3 2 4 1 5 6 3 2 -100000 1 2 100000 1 1 0 2 2 66 3 1 5 4 4 -3
output:
6 100005 6 10 6 10 1 4 1 -1 1 1
result:
ok 6 lines
Test #2:
score: 0
Accepted
time: 0ms
memory: 16084kb
input:
5 6 -10 0 2 -4 8 1 7 1 1 2 2 2 -2 1 1 100 2 1 -100 1 1 0 4 3 10 2 5 3 5 2 2
output:
4 -87 1 17 4 13 1 19 1 17 1 15
result:
ok 6 lines
Test #3:
score: 0
Accepted
time: 0ms
memory: 17912kb
input:
6 3 0 0 0 0 0 0 1 10 1 10 1 -100 4 10 4 11 1 1 0 4 1 0 1 4 1000
output:
2 10 6 11 2 10
result:
ok 3 lines
Test #4:
score: 0
Accepted
time: 0ms
memory: 18012kb
input:
2 0 1 1 1 3
output:
result:
ok 0 lines
Test #5:
score: -100
Memory Limit Exceeded
input:
200000 200000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 ...
output:
119017 15000000000 120167 17000000000 119017 15000000000 119017 15000000000 120167 17000000000 120167 15000000000 120167 16000000000 119017 17000000000 119017 16000000000 119017 12000000000 119017 17000000000 120167 16000000000 120167 14000000000 120167 17000000000 120167 18000000000 120167 16000000...