Submission #120234

#	Time	Username	Problem	Language	Result	Execution time	Memory
120234		turbat	Dreaming (IOI13_dreaming)	C++14	0 / 100	68 ms	15608 KiB

dreaming

#include "dreaming.h"
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
#define ll long long
using pii = pair <int, int>;
using vi = vector <int>;
#define F first
#define S second
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define ook order_of_key
#define fbo find_by_order
#define sq(x) (x) * (x)
#define N 100005
vector <pii> edg[N];
vector <ll> se;
ll dist, dp[N], dp1[N], dp2[N], up[N], up1[N], ans;
int vis[N];
void dfs(int u, ll mx){
	vis[u] = 1;
    if (up[u] < mx){
        up1[u] = up[u];
        up[u] = mx;
    }
    else if (up1[u] < mx) up1[u] = mx;
	for (pii v : edg[u])
		if (!vis[v.F]){
			dfs(v.F, max(mx, dp[u]) + v.S);
            if (dp[u] < dp[v.F] + v.S) {
                dp1[u] = dp[u];
                dp[u] = dp[v.F] + v.S;
            }
            else if (dp1[u] < dp[v.F] + v.S) dp1[u] = dp[v.F] + v.S;
   		}
}
void dfs1(int u, ll mx){
	vis[u] = 2;
    if (up[u] < mx){
        up1[u] = up[u];
        up[u] = mx;
    }
    else if (up1[u] < mx) up1[u] = mx;
	for (int i = edg[u].size() - 1;i >= 0;i--){
		pii v = edg[u][i];
		if (vis[v.F] < 2){
			dfs1(v.F, max(mx, dp2[u]) + v.S);
			dp2[u] = max(dp2[u], dp2[v.F] + v.S);
		}
	}
    ans = max({ans, dp[u] + dp1[u], up[u] + up1[u], dp[u] + up[u]});
	dist = min(dist, max({dp[u], dp1[u], up[u], up1[u]}));
}
int travelTime(int n, int m, int L, int A[], int B[], int T[]) {
	for (int i = 0;i < m;i++){
		edg[A[i]].pb(mp(B[i], T[i]));
		edg[B[i]].pb(mp(A[i], T[i]));
	}
	int k = 0;
	for (int i = 0;i < n;i++)
    	if (!vis[i]){
    		k++;
    		dist = 1e12;
    		dfs(i, 0);
    		dfs1(i, 0);
    		se.pb(dist);
    	}
    sort(se.begin(), se.end());
    if (se.size() == 1) return ans;
	if (se.size() == 2) return max(ans, se[0] + se[1] + 1ll * L);
	return max({se.back() + se[k - 2] + L, se[k - 2] + se[k - 3] + 2ll * L, ans});
}

• Subtask 1: 0 / 14
• Subtask 2: 0 / 10
• Subtask 3: 0 / 23
• Subtask 4: 0 / 18
• Subtask 5: 0 / 12
• Subtask 6: 0 / 23