Submission #954729

#TimeUsernameProblemLanguageResultExecution timeMemory
954729GrindMachineShortcut (IOI16_shortcut)C++17
0 / 100
1 ms436 KiB
#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) (int)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; #include "shortcut.h" long long find_shortcut(int n, std::vector<int> a_, std::vector<int> b_, int c_) { vector<ll> a,b; trav(x,a_) a.pb(x); trav(x,b_) b.pb(x); ll c = c_; a.insert(a.begin(),0); a.pb(0); b.insert(b.begin(),0); vector<ll> p(n+5); rep1(i,n-1){ p[i+1] = p[i]+a[i]; } vector<ll> smx(n+5); rep1(i,n){ smx[i] = max(b[i],a[i-1]+smx[i-1]); } vector<ll> pmx(n+5); rev(i,n,1){ pmx[i] = max(b[i],a[i]+pmx[i+1]); } ll ans = inf2; rep1(i,n){ for(int j = i+1; j <= n; ++j){ ll len = p[j]-p[i]; ll curr = 0; amax(curr,smx[i]+pmx[j]+min(c,len)); ll pref_mx = smx[i-1]+a[i-1]; ll suff_mx = pmx[j+1]+a[j]; for(int k = i; k <= j; ++k){ ll val1 = p[k]-p[i]+b[k]+pref_mx; ll val2 = c+p[j]-p[k]+b[k]+pref_mx; ll mn = min(val1,val2); amax(curr,mn); val1 = p[j]-p[k]+b[k]+suff_mx; val2 = c+p[k]-p[i]+b[k]+suff_mx; mn = min(val1,val2); amax(curr,mn); } for(int x = i; x <= j; ++x){ for(int y = x; y <= j; ++y){ ll connect = min(p[y]-p[x],c+len+p[x]-p[y]); ll val = connect+b[x]+b[y]; if(x == y) val -= b[x]; amax(curr,val); } } amin(ans,curr); } } return ans; }
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