이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "highway.h"
//#include "grader.cpp"
#include <bits/stdc++.h>
#define pii pair<int, int>
#define fr first
#define sc second
#define szof(s) (int)s.size()
#define all(s) s.begin(), s.end()
#define mk make_pair
#define pb push_back
#define ll long long
using namespace std;
const int MAXN = 130005;
const ll INF = 0x3f3f3f3f3f3f3f3f;
int n, m;
vector <pair<int, int>> G[MAXN]; // обычный граф
vector <pair<int, int>> tree[MAXN]; // дерево
int d[MAXN];
int tin[MAXN];
int tout[MAXN];
int tiktak = 0;
int edge[MAXN][2];
bool used[MAXN];
int ROOT, SUB_S;
vector <int> BFS_EDGES;
vector <int> bfs_edges[3]; // already sorted by id
map <pii, int> id;
ll costa, costb;
ll diametr;
ll min_path, max_path;
ll get_min_path() {
vector <int> w(m);
for (int i = 0; i < m; i++) {
w[i] = 0;
}
ll cur_cost = ask(w);
return cur_cost;
}
int find_root() {
vector <int> w(m);
for (int i = 0; i < m; i++) {
w[i] = 0;
}
w[0] = 1;
ll cur_cost;
int l = -1, r = m - 1;
while (r - l > 1) {
int mid = (l + r) >> 1;
for (int i = 0; i < m; i++) {
w[i] = 0;
}
for (int i = 0; i <= mid; i++) {
w[i] = 1;
}
cur_cost = ask(w);
if (cur_cost == min_path) {
l = mid;
} else {
r = mid;
}
}
ROOT = edge[r][0];
SUB_S = edge[r][1];
return edge[r][0];
}
void create_tree(int root) {
BFS_EDGES.clear();
memset(used, 0, sizeof(used));
for (int i = 0; i < MAXN; i++) {
tree[i].clear();
}
// найти ребро на пути S -> T
queue <int> q;
vector <int> dist(MAXN, -1);
dist[root] = 0;
q.push(root);
assert(id.count(mk(root, SUB_S)));
for (int i = 1; i < szof(G[root]); i++) {
if (G[root][i].fr == SUB_S) {
swap(G[root][0], G[root][i]);
break;
}
}
// почему ? надо разобраться
//int sub_id = id[mk(root, SUB_S)];
//used[SUB_S] = true;
//BFS_EDGES.pb(sub_id);
//tree[root].pb({SUB_S, sub_id});
//tree[sub_id].pb({SUB_S, root});
//dist[SUB_S] = 1;
//q.push(SUB_S);
while (!q.empty()) {
int v = q.front();
q.pop();
for (pii to : G[v]) {
if (dist[to.fr] == -1) {
used[to.sc] = true;
BFS_EDGES.pb(to.sc);
tree[v].pb({to.fr, to.sc});
tree[to.fr].pb({v, to.sc});
dist[to.fr] = dist[v] + 1;
q.push(to.fr);
}
}
}
}
void calc(int v, int par, int len) {
tin[v] = ++tiktak;
d[v] = len;
for (pii to : tree[v]) {
if (to.fr != par) {
calc(to.fr, v, len + 1);
}
}
tout[v] = tiktak;
}
bool father(int a, int b) {
return tin[a] <= tin[b] && tout[b] <= tout[a];
}
vector <int> subtree[3];
void dfs(int v, int par) {
if (father(SUB_S, v)) {
subtree[1].pb(v);
} else {
subtree[2].pb(v);
}
for (pii to : tree[v]) {
if (to.fr != par) {
dfs(to.fr, v);
}
}
}
int find_ans(int cur_root, vector <int> cur_bfs_edges) {
if (cur_bfs_edges.empty()) {
return cur_root;
}
int l = -1, r = szof(cur_bfs_edges);
vector <int> w(m);
while (r - l > 1) {
int mid = (l + r) >> 1;
for (int i = 0; i < m; i++) {
w[i] = 1;
}
for (int el : BFS_EDGES) {
w[el] = 0;
}
for (int i = mid; i < szof(cur_bfs_edges); i++) {
w[cur_bfs_edges[i]] = 1;
}
ll cur_cost = ask(w);
if (cur_cost == min_path) {
r = mid;
} else {
l = mid;
}
}
if (l == -1) {
return cur_root;
}
int x = edge[cur_bfs_edges[l]][0];
int y = edge[cur_bfs_edges[l]][1];
return (d[x] > d[y] ? x : y);
}
void find_pair(int N, vector<int> U, vector<int> V, int AA, int BB) {
m = szof(U);
n = N;
costa = AA;
costb = BB;
min_path = get_min_path(); // 1 q
diametr = min_path / costa;
max_path = diametr * costb;
for (int i = 0; i < m; i++) {
int x, y;
x = U[i];
y = V[i];
id[{x, y}] = i;
id[{y, x}] = i;
edge[i][0] = x;
edge[i][1] = y;
G[x].pb({y, i});
G[y].pb({x, i});
}
int s, t;
int root = find_root(); // log(n)
create_tree(root);
calc(root, -1, 0);
dfs(root, -1);
sort(all(subtree[1]));
sort(all(subtree[2]));
auto exist = [&](vector <int> &vec, int x) {
bool ok = binary_search(all(vec), x);
return ok;
};
for (int el : BFS_EDGES) {
int x = edge[el][0];
int y = edge[el][1];
if ((x == ROOT && y == SUB_S) || (x == SUB_S && y == ROOT)) {
continue;
}
if (exist(subtree[1], x)) {
bfs_edges[1].pb({id[{x, y}]});
} else {
bfs_edges[2].pb({id[{x, y}]});
}
}
s = find_ans(SUB_S, bfs_edges[1]); // log(n) - 1
t = find_ans(ROOT, bfs_edges[2]); // log(n) - 1
answer(s, t);
}
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