Submission #955082

#TimeUsernameProblemLanguageResultExecution timeMemory
955082GrindMachineMiners (IOI07_miners)C++17
100 / 100
221 ms201804 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; void solve(int test_case) { ll n; cin >> n; string s; cin >> s; s = "$" + s; vector<ll> a(n+5); rep1(i,n){ if(s[i] == 'B') a[i] = 1; if(s[i] == 'F') a[i] = 2; if(s[i] == 'M') a[i] = 3; } ll dp[n+5][4][4][4][4]; memset(dp,-0x3f,sizeof dp); dp[1][0][0][0][0] = 0; rep1(i,n){ rep(l1,4){ rep(l2,4){ rep(l3,4){ rep(l4,4){ ll curr = dp[i][l1][l2][l3][l4]; // mine 1 { ll add = 1; if(l2 and l2 != a[i]) add++; if(l1 and l1 != l2 and l1 != a[i]) add++; amax(dp[i+1][l2][a[i]][l3][l4],curr+add); } // mine 2 { ll add = 1; if(l4 and l4 != a[i]) add++; if(l3 and l3 != l4 and l3 != a[i]) add++; amax(dp[i+1][l1][l2][l4][a[i]],curr+add); } } } } } } ll ans = 0; rep(l1,4){ rep(l2,4){ rep(l3,4){ rep(l4,4){ amax(ans,dp[n+1][l1][l2][l3][l4]); } } } } cout << ans << endl; } int main() { fastio; int t = 1; // cin >> t; rep1(i, t) { solve(i); } return 0; }
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