Submission #860340

#	Time	Username	Problem	Language	Result	Execution time	Memory
860340		wii	Popeala (CEOI16_popeala)	C++17	100 / 100	1664 ms	44352 KiB

popeala

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

using namespace __gnu_pbds;
using namespace std;

typedef double db;
typedef long long ll;
typedef long double ld;

#define int ll
typedef pair<int, int> pii;

#define X first
#define Y second
#define gcd __gcd
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define bit(i, mask) ((mask) >> (i) & 1)
#define reset(x, val) memset(x, val, sizeof(x))
#define foru(i,a,b) for(int i = (a); i <= (b); ++i)
#define ford(i,a,b) for(int i = (a); i >= (b); --i)
#define FastIO ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

template<typename T> bool maximize(T &res, const T &val) { if (res < val) { res = val; return true; } return false; }
template<typename T> bool minimize(T &res, const T &val) { if (val < res) { res = val; return true; } return false; }

const ll Linf = 0x3f3f3f3f3f3f3f3f;
const int Inf = 0x3f3f3f3f;
const ll Mod = 1e9 + 7;
const ll Mod2 = ll(1e9) + 9;
const int Lim = 1e6 + 5;
const int inv6 = 166666668;

// #define Sieve
#ifdef Sieve
    vector<int> pr;
    vector<int> lpf;
    void Linear_sieve(int n = Lim) {
        pr.assign(1, 2);
        lpf.assign(n + 1, 2);

        for (int x = 3; x <= n; x += 2) {
            if (lpf[x] == 2) pr.push_back(lpf[x] = x);
            for (int i = 1; i < pr.size() && pr[i] <= lpf[x] && pr[i] * x <= n; ++i)
                lpf[pr[i] * x] = pr[i];
        }
    }
#endif

// #define Ckn_calc
#ifdef Ckn_calc
    const int LIM = 1e6 + 16;

    int fact[LIM + 10]; /// factorial:         fact[n] = n!
    int invs[LIM + 10]; /// inverse modular:   invs[n] = n^(-1)
    int tcaf[LIM + 10]; /// inverse factorial: tcaf[n] = (n!)^(-1)
    void precal_nck(int n = LIM)
    {
        fact[0] = fact[1] = 1;
        invs[0] = invs[1] = 1;
        tcaf[0] = tcaf[1] = 1;
        for (int i = 2; i <= n; ++i)
        {
            fact[i] = (1LL * fact[i - 1] * i) % Mod;
            invs[i] = Mod - 1LL * (Mod / i) * invs[Mod % i] % Mod;
            tcaf[i] = (1LL * tcaf[i - 1] * invs[i]) % Mod;
        }
    }

    ll C(int n, int k) {
        if (n < k || k < 0) return 0;

        ll res = fact[n];
        res *= tcaf[k], res %= Mod;
        res *= tcaf[n - k], res %= Mod;
        return res;
    }

    ll P(int n, int k) {
        ll res = fact[n];
        res *= tcaf[n - k], res %= Mod;
        return res;
    }
#endif

/// ====*====*====*====*====*====*====*====*====*====*====*====*====*====*====*====*====

const int base = 3;
const int N = 2e4 + 5;
const int K = 50 + 5;
const int dx[] = {+1, -1, 0, 0};
const int dy[] = {0, 0, +1, -1};
const int block_size = sqrt(2e9) + 2;

int n, t, k;
int a[N], last[N][K], dp[N][K], f[N][K];
int opt[N], cnt[N];

int full_row(int l, int r) {
    int ans = 0;
    foru(i, 1, n)
        ans += (f[r][i] - f[l - 1][i] == r - l + 1);
    return ans;
}

int seg[K][N * 2];
void update(int mul, int u, int x) {
    for (seg[mul][u += t] = x; u > 1; u >>= 1)
        seg[mul][u >> 1] = min(seg[mul][u], seg[mul][u ^ 1]);
}

int get(int mul, int l, int r) {
    int ans = Linf;
    for (l += t, r += t + 1; l < r; l >>= 1, r >>= 1) {
        if (l & 1) minimize(ans, seg[mul][l++]);
        if (r & 1) minimize(ans, seg[mul][--r]);
    }
    return ans;
}

void solve() {
    cin >> n >> t >> k;
    foru(i, 1, t) cin >> a[i], a[i] += a[i - 1];

    foru(i, 1, n) {
        char x;
        foru(j, 1, t) {
            cin >> x;
            last[j][i] = (x == '0' ? j : last[j - 1][i]);
            f[j][i] = f[j - 1][i] + (x == '1');
        }
    }

    foru(j, 1, t) sort(last[j] + 1, last[j] + n + 1);

    memset(dp, 0x3f, sizeof dp);
    dp[0][0] = 0;

    foru(z, 1, k) {
        foru(x, 0, n) {
            foru(i, 1, t) {
                opt[i] = dp[i - 1][z - 1] - x * a[i - 1];
                update(x, i, opt[i]);
            }
        }
        
        foru(i, 1, t) {
            foru(p, 1, n)
                cnt[last[i][p]] += 1;
            last[i][n + 1] = i;

            foru(p, 1, n + 1) if (last[i][p] != last[i][p - 1])
                cnt[last[i][p]] += cnt[last[i][p - 1]];

            foru(p, 1, n + 1) {
                int l = last[i][p - 1] + 1, r = last[i][p];
                int x = cnt[l - 1];

                int ans = get(x, l, r);
                minimize(dp[i][z], ans + a[i] * x);
            }

            foru(p, 0, n + 1) cnt[last[i][p]] = 0;
        }

        cout << dp[t][z] << "\n";
    }
}

signed main() {
    FastIO;

    #define task "test"
    if (fopen(task".inp", "r")) {
		freopen(task".inp", "r", stdin);
		freopen(task".out", "w", stdout);
	}

    #ifdef Sieve
        Linear_sieve();
    #endif

    #ifdef Ckn_calc
        precal_nck();
    #endif

    int ntest = 1;
    // cin >> ntest;
    while (ntest--) {
        //cout << "Case " << q << ": " << "\n"; 
        solve();
        cout << "\n";
    }

    return 0;
}

/**  /\_/\
 *  (= ._.)
 *  / >TL \>AC
**/

Compilation message (stderr)

popeala.cpp: In function 'int main()':
popeala.cpp:176:10: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  176 |   freopen(task".inp", "r", stdin);
      |   ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
popeala.cpp:177:10: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  177 |   freopen(task".out", "w", stdout);
      |   ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 8 / 8
• Subtask 2: 9 / 9
• Subtask 3: 9 / 9
• Subtask 4: 74 / 74