제출 #1199412

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1199412		SSSM	007 (CEOI14_007)	C++20	100 / 100	216 ms	48980 KiB

007

#include <bits/stdc++.h>

/*
#pragma GCC optimize("O3,unroll-loops")
#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 FIO freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout)
#define md(a) (a%mod+mod)%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=6e5+10, mod=1e9+7, INF=1e9, LOG=20, 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, k, t, a, b, dis[4][maxn];
vector<ll> g[maxn];

void BFS(ll ind, ll s)
{
    fill(dis[ind], dis[ind]+maxn, INF);
    dis[ind][s]=0;
    queue<ll> q;
    q.push(s);
    while(SZ(q))
    {
        ll v=q.front();
        q.pop();
        for(ll u:g[v])
        {
            if(dis[ind][u]>dis[ind][v]+1)
            {
                dis[ind][u]=dis[ind][v]+1;
                q.push(u);
            }
        }
    }

}

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

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

    BFS(0, a);
    BFS(1, b);
    BFS(2, k);
    BFS(3, t);

    ll K=INF, T=INF;
    if(dis[0][k]==dis[1][k])
    {
        for(ll v=1;v<=n;v++) if(dis[0][v]==dis[1][v] && dis[0][v]+dis[2][v]==dis[0][k])
            K=min(K, dis[0][v]);
    }
    if(dis[0][t]==dis[1][t])
    {
        for(ll v=1;v<=n;v++) if(dis[0][v]==dis[1][v] && dis[0][v]+dis[3][v]==dis[0][t])
            T=min(T, dis[0][v]);
    }    

    ll dk=min(dis[0][k], dis[1][k]);
    ll dt=min(dis[0][t], dis[1][t]);
    ll ans=(dt-dk);
    if(K==INF && T==INF)
    {
        bool ok=dis[0][k]<dis[1][k];
        bool ot=dis[0][t]<dis[1][t];
        if(ok!=ot) ans--;
    }
    else ans-=(K>T);
    if(ans<0) ans=-1;
    cout<<ans<<"\n";
}

• Subtask 1: 30 / 30
• Subtask 2: 70 / 70