이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
using namespace std;
#define int long long
#define vi vector<int>
#define vl vector<long long>
#define vii vector<pair<int,int>>
#define vll vector<pair<long long,long long>>
#define pb push_back
#define ll long long
#define ld long double
#define nl '\n'
#define boost ios::sync_with_stdio(false)
#define mp make_pair
#define se second
#define fi first
#define fore(i, y) for(int i = 0; i < y; i++)
#define forr(i,x,y) for(int i = x;i<=y;i++)
#define forn(i,y,x) for(int i = y; i >= x; i--)
#define all(v) v.begin(),v.end()
#define sz(v) v.size()
#define clr(v,k) memset(v,k,sizeof(v))
#define rall(v) v.rbegin() , v.rend()
#define pii pair<int,int>
#define pll pair<ll , ll>
const ll MOD = 1e9 + 7;
const int INF = 1e18 + 1;
const int inf = 1e9 + 1;
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (gcd)
ll lcm(ll a , ll b) {return a * (b / gcd(a , b));} // least common multiple (lcm)
// HERE IS THE SOLUTION
const int N = 100000;
vii adj[N];
int n , m , s , t , u , v;
vi dijkstra(int start)
{
vi dist(n , INF);
priority_queue<pii> pq;
pq.push({0 , start});
dist[start] = 0;
vector<bool> vis(n , 0);
while(!pq.empty())
{
int cur = pq.top().se;
pq.pop();
if(vis[cur])
continue;
vis[cur] = 1;
for(auto [x , c] : adj[cur])
{
if(dist[x] > dist[cur] + c){
dist[x] = dist[cur] + c;
pq.push({-dist[x] , x});
}
}
}
return dist;
}
signed main()
{
cin>>n>>m>>s>>t>>u>>v;
if(n > 500)
{
cout<<0<<nl;
return 0;
}
u-- , v-- , t-- , s--;
vector<vi> dist(n,vi(n , INF));
fore(i , n)
dist[i][i] = 0;
fore(i , m)
{
int a , b,c;
cin>>a>>b>>c;
a-- , b--;
dist[a][b] = dist[b][a] = c;
adj[a].pb({b , c});
adj[b].pb({a , c});
}
fore(k , n)
{
fore(i , n)
{
fore(j ,n)
{
dist[i][j] = dist[j][i] = min(dist[i][k] + dist[k][j] , dist[i][j]);
}
}
}
vi spS = dijkstra(s) , spT = dijkstra(t) , spV = dijkstra(v) , spU = dijkstra(u);
int ans = INF;
fore(ai , n)
{
fore(aj , n)
{
if(dist[s][ai] + dist[ai][aj] + dist[aj][t] == dist[s][t])
{
ans = min({ans , dist[u][ai] + dist[aj][v] , dist[u][aj] + dist[ai][v]});
}
}
}
cout<<min(ans , dist[u][v])<<nl;
}
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |