Submission #739207

#TimeUsernameProblemLanguageResultExecution timeMemory
739207GrindMachinePopeala (CEOI16_popeala)C++17
26 / 100
2064 ms8404 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, t, s; cin >> n >> t >> s; vector<ll> a(t + 5); rep1(i, t) cin >> a[i]; vector<ll> p(t + 5); rep1(i, t) p[i] = p[i - 1] + a[i]; ll b[n + 5][t + 5]; rep1(i, n) { rep1(j, t) { char ch; cin >> ch; b[i][j] = ch - '0'; } } ll prev_bad[n + 5][t + 5]; memset(prev_bad, -1, sizeof prev_bad); rep1(i, n) { rep1(j, t) { if (!b[i][j]) { prev_bad[i][j] = j; } else { prev_bad[i][j] = prev_bad[i][j - 1]; } } } vector<ll> dp1(t + 5, inf2), dp2(t + 5, inf2); dp1[0] = 0; vector<ll> change(t + 5); rep1(iter, s) { rep1(i, t) { fill(all(change), 0); rep1(j, n) { ll pos = prev_bad[j][i]; if (pos != -1) { change[pos]++; } } ll g = n; rev(k, i, 1) { g -= change[k]; ll val = dp1[k - 1] + g * (p[i] - p[k - 1]); amin(dp2[i], val); } } ll ans = dp2[t]; cout << ans << endl; dp1 = dp2; fill(all(dp2), inf2); } } int main() { fastio; int t = 1; // cin >> t; rep1(i, t) { solve(i); } return 0; }
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