이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#pragma GCC optimize("Ofast,unroll-loops")
#include <bits/stdc++.h>
#define ll long long
#define pii pair<int, int>
#define pll pair<ll, ll>
#define F first
#define S second
using namespace std;
const int B = 550;
int N, M, Q, u[100005], v[100005], d[100005], nd[100005], ans[100005], dsu[50005], rk[50005], vis[100005];
vector<pii> ops;
int find(int k)
{
return k == dsu[k] ? k : find(dsu[k]);
}
void merge(int x, int y)
{
x = find(x), y = find(y);
if(x == y)
return;
if(rk[x] < rk[y])
merge(y, x);
else
{
dsu[y] = x;
rk[x] += rk[y];
ops.emplace_back(pii(x, y));
}
}
void undo(int t)
{
while((int)ops.size() > t)
{
auto [x, y] = ops.back();
ops.pop_back();
dsu[y] = y, rk[x] -= rk[y];
}
}
void init()
{
for (int i = 1; i <= N; i++)
dsu[i] = i, rk[i] = 1;
}
signed main()
{
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
cin >> N >> M;
for (int i = 1; i <= M; i++)
cin >> u[i] >> v[i] >> d[i];
cin >> Q;
for (int i = 1; i <= Q; i += B)
{
init();
for (int j = 1; j <= M; j++)
vis[j] = 0;
// solve Q = [i, i + B - 1]
vector<tuple<int, int, int>> change;
vector<int> out;
// d, i, 0: addedge
// d, i, q: query i (ans[q])
vector<tuple<int, int, int>> event;
for (int j = i; j <= min(Q, i + B - 1); j++)
{
int t, a, b;
cin >> t >> a >> b;
if(t == 1)
{
change.emplace_back(make_tuple(j, a, b));
vis[a] = 1;
}
else
event.emplace_back(make_tuple(-b, j, a));
}
for (int j = 1; j <= M; j++)
if(!vis[j])
event.emplace_back(make_tuple(-d[j], 0, j));
else
{
change.emplace_back(make_tuple(i - 1, j, d[j]));
out.emplace_back(j);
}
sort(event.begin(), event.end());
sort(change.begin(), change.end());
for (auto [d, q, j] : event)
{
//cerr << "event " << d << ' ' << t << ' ' << j << ' ' << q << '\n';
if(q == 0)
{
//cerr << "include " << j << '\n';
merge(u[j], v[j]);
}
else
{
int T = ops.size();
for (auto [tt, k, dd] : change)
if(tt <= q)
nd[k] = dd;
else break;
for (auto k : out)
if(nd[k] >= -d)
{
//cerr << "tmp include " << k << '\n';
merge(u[k], v[k]);
}
ans[q] = rk[find(j)];
undo(T);
}
}
for (auto [t, j, dd] : change)
d[j] = dd;
}
for (int i = 1; i <= Q; i++)
{
if(ans[i] > 0)
cout << ans[i] << '\n';
}
}
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