Submission #123659

#	Time	Username	Problem	Language	Result	Execution time	Memory
123659		eriksuenderhauf	Highway Tolls (IOI18_highway)	C++11	51 / 100	3065 ms	14096 KiB

highway

#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include "highway.h"
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
#define mem(a,v) memset((a), (v), sizeof (a))
#define enl printf("\n")
#define case(t) printf("Case #%d: ", (t))
#define ni(n) scanf("%d", &(n))
#define nl(n) scanf("%I64d", &(n))
#define nai(a, n) for (int i = 0; i < (n); i++) ni(a[i])
#define nal(a, n) for (int i = 0; i < (n); i++) nl(a[i])
#define pri(n) printf("%d\n", (n))
#define prl(n) printf("%I64d\n", (n))
#define pii pair<int, int>
#define pil pair<int, long long>
#define pll pair<long long, long long>
#define vii vector<pii>
#define vil vector<pil>
#define vll vector<pll>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef cc_hash_table<int,int,hash<int>> ht;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> oset;
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 130050;
const double eps = 1e-9;
vii adj[MAXN], all;
int n, m, vis[MAXN], dp[2][MAXN];
ll dist;

int solve(int root, int idx) {
	deque<int> pq;
	pq.pb(root);
	vi edg, tmp(m, 1);
	while (!pq.empty()) {
		int u = pq.front(); pq.pop_front();
		for (pii nx: adj[u]) {
			if (dp[idx][nx.fi] != dp[idx][u] + 1) continue;
			tmp[nx.se] = 0;
			pq.pb(nx.fi);
			if (dp[idx][nx.fi] < dp[idx^1][nx.fi])
				edg.pb(nx.se);
		}
	}
	reverse(edg.begin(), edg.end());
	int lo = 0, hi = edg.size()-1, ans = -1;
	while (lo <= hi) {
		int mi = (lo+hi) / 2;
		vi cur;
		for (int i = 0; i <= mi; i++)
			tmp[edg[i]] = 1;
		ll w = ask(tmp);
		if (w == dist)
			lo = mi + 1;
		else {
			ans = mi;
			hi = mi - 1;
		}
		for (int i = 0; i <= mi; i++)
			tmp[edg[i]] = 0;
	}
	if (ans == -1) 
		return root;	
	pii ret = all[edg[ans]];
	if (dp[idx][ret.fi] < dp[idx][ret.se])
		swap(ret.fi, ret.se);
	return ret.fi;
}

void find_pair(int N, vi U, vi V, int a, int b) {
	n = N;
	m = U.size();
	for (int i = 0; i < m; i++) {
		adj[U[i]].pb(mp(V[i],i));
		adj[V[i]].pb(mp(U[i],i));
		all.pb(mp(U[i], V[i]));
	}
	dist = ask(vi(m, 0));
	int lo = 0, hi = m-1;
	while (lo <= hi) {
		int mi = (lo+hi) / 2;
		vi tmp(m);
		for (int i = 0; i <= mi; i++)
			tmp[i] = 1;
		ll w = ask(tmp);
		if (w == dist)
			lo = mi + 1;
		else
			hi = mi - 1;
	}
	int u = all[lo].fi, v = all[lo].se;
	mem(vis, 0);
	mem(dp, 0);
	deque<int> pq;
	vis[u] = 1; pq.pb(u);
	while (!pq.empty()) {
		int u2 = pq.front(); pq.pop_front();
		for (pii nx: adj[u2]) {
			if (vis[nx.fi]) continue;
			vis[nx.fi] = 1;
			dp[0][nx.fi] = dp[0][u2] + 1;
			pq.pb(nx.fi);
		}
	}
	mem(vis, 0);
	vis[v] = 1; pq.pb(v);
	while (!pq.empty()) {
		int u2 = pq.front(); pq.pop_front();
		for (pii nx: adj[u2]) {
			if (vis[nx.fi]) continue;
			vis[nx.fi] = 1;
			dp[1][nx.fi] = dp[1][u2] + 1;
			pq.pb(nx.fi);
		}
	}
	int s = solve(u, 0);
	int t = solve(v, 1);
	answer(s, t);
}

• Subtask 1: 5 / 5
• Subtask 2: 7 / 7
• Subtask 3: 6 / 6
• Subtask 4: 33 / 33
• Subtask 5: 0 / 18
• Subtask 6: 0 / 31