Submission #759513

#TimeUsernameProblemLanguageResultExecution timeMemory
759513GrindMachinePinball (JOI14_pinball)C++17
51 / 100
1085 ms132676 KiB
// Om Namah Shivaya #include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; typedef long long int ll; typedef long double ld; typedef pair<int, int> pii; typedef pair<ll, ll> pll; #define fastio ios_base::sync_with_stdio(false); cin.tie(NULL) #define pb push_back #define endl '\n' #define sz(a) a.size() #define setbits(x) __builtin_popcountll(x) #define ff first #define ss second #define conts continue #define ceil2(x, y) ((x + y - 1) / (y)) #define all(a) a.begin(), a.end() #define rall(a) a.rbegin(), a.rend() #define yes cout << "Yes" << endl #define no cout << "No" << endl #define rep(i, n) for(int i = 0; i < n; ++i) #define rep1(i, n) for(int i = 1; i <= n; ++i) #define rev(i, s, e) for(int i = s; i >= e; --i) #define trav(i, a) for(auto &i : a) template<typename T> void amin(T &a, T b) { a = min(a, b); } template<typename T> void amax(T &a, T b) { a = max(a, b); } #ifdef LOCAL #include "debug.h" #else #define debug(x) 42 #endif /* */ const int MOD = 1e9 + 7; const int N = 1e5 + 5; const int inf1 = int(1e9) + 5; const ll inf2 = ll(1e18) + 5; void solve(int test_case) { ll n,m; cin >> n >> m; vector<array<ll,4>> a(n+5); for(int i = 3; i < n+3; ++i){ rep(j,4){ cin >> a[i][j]; } } a[1] = {1,1,1,0}; a[2] = {m,m,m,0}; auto dijkstra = [&](ll src, vector<ll> &dis){ priority_queue<pll,vector<pll>,greater<pll>> pq; pq.push({0,src}); vector<bool> vis(n+5); while(!pq.empty()){ auto [cost,i] = pq.top(); pq.pop(); if(vis[i]) conts; vis[i] = 1; dis[i] = cost; for(int j = i+1; j <= n+2; ++j){ auto [l1,r1,k1,w1] = a[i]; auto [l2,r2,k2,w2] = a[j]; if(l2 <= k1 and k1 <= r2){ pq.push({cost+w2,j}); } } } }; vector<ll> dis1(n+5,inf2), dis2(n+5,inf2); dijkstra(1,dis1); dijkstra(2,dis2); ll ans = inf2; rep1(i,n+2){ ll cost = dis1[i] + dis2[i] - a[i][3]; amin(ans,cost); } if(ans == inf2) ans = -1; cout << ans << endl; } int main() { fastio; int t = 1; // cin >> t; rep1(i, t) { solve(i); } return 0; }
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