제출 #307524

#제출 시각아이디문제언어결과실행 시간메모리
307524shivensinha4Commuter Pass (JOI18_commuter_pass)C++17
0 / 100
1 ms512 KiB
#include <bits/stdc++.h> using namespace std; #define for_(i, s, e) for (int i = s; i < (int) e; i++) #define for__(i, s, e) for (ll i = s; i < e; i++) typedef long long ll; typedef vector<int> vi; typedef pair<int, int> ii; #define endl '\n' const int MXN = 10; const ll INF = 1e15; int n, m; vector<pair<int, ll>> adj[MXN+1]; //ll du[MXN+1], dv[MXN+1]; //pair<ll, ll> df[MXN+1][3][3]; //void dijk(int a, ll d[]) { //for_(i, 0, n) d[i] = INF; //d[a] = 0; //priority_queue<pair<ll, int>, deque<pair<ll, int>>, greater<pair<ll, int>>> pq; //pq.push({0, a}); //while (pq.size()) { //int p = pq.top().first; ll dist = pq.top().second; pq.pop(); //if (d[p] < dist) continue; //for (auto i: adj[p]) if (d[i.first] > dist+i.second) { //d[i.first] = dist+i.second; //pq.push({d[i.first], i.first}); //} //} //} int main() { #ifdef shiven freopen("test.in", "r", stdin); #endif ios_base::sync_with_stdio(false); cin.tie(0); int s, t, u, v; cin >> n >> m >> s >> t >> u >> v; s -= 1; t -= 1; u -= 1; v -= 1; for_(i, 0, m) { int a, b; ll w; cin >> a >> b >> w; a -= 1; b -= 1; adj[a].push_back({b, w}); adj[b].push_back({a, w}); } //dijk(u, du); dijk(v, dv); //for_(i, 0, n) { //adj[i].push_back({i, 0}); //for_(j, 0, 2) for_(k, 0, 2) df[i][j][k] = {INF, INF}; //} //priority_queue<pair<pair<ll, ll>, vi>, deque<pair<pair<ll, ll>, vi>>, greater<pair<pair<ll, ll>, vi>>> pq; //pq.push({{0, 0}, {s, 0, 0}}); //ll ans = INF, bda = INF; //while (pq.size()) { //auto p = pq.top().second; //auto da = pq.top().first.first, db = pq.top().first.second; pq.pop(); //if (df[p[0]][p[1]][p[2]] < pair<ll, ll> {da, db}) continue; //if (p == (vi) {t, 1, 1}) { //if (da < bda) ans = db; //else if (da == bda) ans = min(ans, db); //bda = min(bda, da); //} //for (auto i: adj[p[0]]) { //if (df[i.first][p[1]][p[2]] > pair<ll, ll> {da+i.second, db}) { //df[i.first][p[1]][p[2]] = {da+i.second, db}; //pq.push({{da+i.second, db}, {i.first, p[1], p[2]}}); //} //if (!p[1] and df[i.first][1][p[2]] > pair<ll, ll> {da+i.second, db+du[p[0]]}) { //df[i.first][1][p[2]] = {da+i.second, db+du[p[0]]}; //pq.push({{da+i.second, db+du[p[0]]}, {i.first, 1, p[2]}}); //} //if (!p[2] and df[i.first][p[1]][1] > pair<ll, ll> {da+i.second, db+dv[p[0]]}) { //df[i.first][p[2]][1] = {da+i.second, db+dv[p[0]]}; //pq.push({{da+i.second, db+dv[p[0]]}, {i.first, p[1], 1}}); //} //} //} cout << 2 << endl; return 0; }
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