Submission #1149139

#	Time	Username	Problem	Language	Result	Execution time	Memory
1149139		SSSM	One-Way Streets (CEOI17_oneway)	C++20	0 / 100	2 ms	4928 KiB

oneway

#include <bits/stdc++.h>

/*
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC target ("avx2")
*/

using namespace std;

/*
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds;
template<class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
*/

#define F first
#define S second
#define pb push_back
#define md(a) (a%mod)
#define all(a) a.begin(), a.end()
#define MP make_pair
#define lc (id<<1)
#define rc (lc|1)
#define mid (l+r)/2
#define kill(a) cout << a << "\n", exit(0)
#define SZ(a) (ll)a.size()

typedef pair<int,int> pii;
typedef pair<long long ,long long> pll;
typedef long long ll;
typedef long double ld;
typedef vector<vector<ll>> matrix;
mt19937_64  rng(chrono::steady_clock::now().time_since_epoch().count());

ll const maxn=2e5+10, mod=1e9+7, INF=1e18, LOG=31, sq=65;

ll poww(ll a, ll b, ll mod) {
 if (b == 0) return 1;
 return 1 * poww(1 * a * a % mod, b / 2, mod) * ((b % 2 == 1) ? a : 1) % mod;
}

ll n, m, dp[maxn], dp2[maxn];
vector<pll> g[maxn];
char ans[maxn];
bool vis[maxn], Vvis[maxn];

void DFS(ll v)
{
    Vvis[v]=1;
    for(auto [u, i]:g[v])
    {
        if(vis[abs(i)]) continue;
        vis[abs(i)]=1;
        if(!Vvis[u])
        {
            DFS(u);
            if(dp2[u]>0||dp[u]==0) ans[abs(i)]='B';
            else ans[abs(i)]=((dp[u]<0)^(i>0))?'L':'R';
            dp[v]+=dp[u];
            dp2[v]+=dp2[u];
        }
        else{
            ans[abs(i)]='B';
            dp2[v]++;
            dp2[u]--;
        }
    }
}

int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

    cin>>n>>m;
    for(ll i=1;i<=m;i++)
    {
        ll v, u;
        cin>>v>>u;
        g[v].pb({u, i});
        g[u].pb({v, -i});
    }

    ll q;
    cin>>q;
    while(q--)
    {
        ll v, u;
        cin>>v>>u;
        dp[v]++;
        dp[u]--;
    }
    DFS(1);
    for(ll i=1;i<=m;i++) cout<<ans[i];
    cout<<"\n";


    return 0;
}

• Subtask 1: 0 / 30
• Subtask 2: 0 / 30
• Subtask 3: 0 / 40