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;
struct tree {
vector<vector<array<long long, 2>>> adj;
void dfs(int ba, vector<long long> &dist, vector<int> &p) {
for (auto x : adj[ba]) {
if (x[0] == p[ba]) continue;
dist[x[0]] = dist[ba]+x[1];
p[x[0]] = ba;
dfs(x[0], dist, p);
}
}
int m(int N, int X, int Y, long long K, vector<int> &U, vector<int> &V, vector<int> &W) {
vector<long long> dx(N, 0);
vector<long long> dy(N, 0);
vector<int> px(N, -1);
vector<int> py(N, -1);
adj.resize(N);
for (long long i = 0; i < N-1; i++ ) {
adj[U[i]].push_back({V[i], W[i]});
adj[V[i]].push_back({U[i], W[i]});
}
dfs(X, dx, px);
dfs(Y, dy, py);
vector<int> acx(N, 0); vector<int> acy(N, 0); // ho gia' comprato la x,y
vector<int> rex(N, 0); vector<int> rey(N, 0); //
priority_queue<array<long long, 3>> d;
for (long long i = 0; i < N; i++) {
if (i == X) {
acx[X] = 1; rex[X] = 1;
for (auto x : adj[X]) {
if (x[0] == px[X]) continue;
rex[x[0]] = 1;
}
}
else if (i == Y) {
acy[Y] = 1; rey[Y] = 1;
for (auto x : adj[Y]) {
if (x[0] == py[Y]) continue;
rey[x[0]] = 1;
}
}
else if (dx[i] < dy[i]) d.push({-dx[i], i, 0});
else d.push({-dy[i], i, 1});
}
vector<int> done(N, 0);
long long att = 0, k = 2;
while (k < 2*N && att-d.top()[0] <= K) {
att -= d.top()[0]; k++;
auto a = d.top(); d.pop();
if (a[2]) {
acy[a[1]] = 1;
if (!done[a[1]] && rex[a[1]] == 1) {
done[a[1]] =1 ;
d.push({-abs(dx[a[1]]-dy[a[1]]), a[1], 0});
}
for (auto x : adj[a[1]]) {
if (x[0] == py[a[1]]) continue;
rey[x[0]] = 1;
if (!done[x[0]] && acx[x[0]] == 1) {
done[x[0]] = 1;
d.push({-abs(dx[x[0]]-dy[x[0]]), x[0], 1});
}
}
}
else {
acx[a[1]] = 1;
if (!done[a[1]] && rey[a[1]] == 1) {
done[a[1]] = 1;
d.push({-abs(dx[a[1]]-dy[a[1]]), a[1], 1});
}
for (auto x : adj[a[1]]) {
if (x[0] == px[a[1]]) continue;
rex[x[0]] = 1;
if (!done[x[0]] && acy[x[0]] == 1) {
done[x[0]] = 1;
d.push({-abs(dx[x[0]]-dy[x[0]]), x[0], 0});
}
}
}
if (a[1] == -1) continue;
}
return k;
}
};
int max_score(int N, int X, int Y, long long K,vector<int> U, vector<int> V, vector<int> W) {
tree a;
return a.m(N, X, Y, K, U, V, W);
}
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