답안 #821352

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
821352 2023-08-11T09:27:16 Z Shreyan_Paliwal Nafta (COI15_nafta) C++17
100 / 100
334 ms 123656 KB
// #pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;

template<typename T>
string tostr(const T& value) {
    ostringstream oss;
    oss << value;
    return oss.str();
}

template<typename... Args>
string fstr(const string& format, Args... args) {
    string result = format;
    size_t pos = 0;
    size_t argIndex = 0;

    auto replaceArg = [&](const auto& arg) {
        pos = result.find("{}", pos);
        if (pos != string::npos) {
            result.replace(pos, 2, tostr(arg));
            ++argIndex;
        }
    };

    (replaceArg(args), ...);

    return result;
}

/*
 * Keeps mint objects modulo MOD constant
*/

// const int MOD = 1e9 + 7;
// struct mint {
//     int x;

//     mint() { x = 0; }
//     mint(int X) { x = X; }
//     mint(long long X) { x = ((X % MOD) + MOD) % MOD; }
//     mint(unsigned long long X) { x = (X % MOD); }

//     mint pow(int k) { mint r = 1, a = *this; while (k) { if (k & 1) r *= a; a *= a; k >>= 1; } return r; }

//     mint& operator+=(mint o) { if ((x += o.x) >= MOD) x -= MOD; return *this; }
//     mint& operator-=(mint o) { if ((x += MOD - o.x) >= MOD) x -= MOD; return *this; }
//     mint& operator*=(mint o) { x = 1ll * x * o.x % MOD; return *this; }
//     mint& operator/=(mint o) { return (*this) *= o.pow(MOD - 2); }

//     mint operator+(mint o) const { return mint(*this) += o; }
//     mint operator-(mint o) const { return mint(*this) -= o; }
//     mint operator*(mint o) const { return mint(*this) *= o; }
//     mint operator/(mint o) const { return mint(*this) /= o; }

//     bool operator<(mint o) const { return x < o.x; }

//     friend ostream& operator<<(ostream& os, mint a) { os << a.x; return os; }
// };

/*
 * Matrices
 * Multiplication
*/

// typedef vector<vector<int>> MAT;
// struct Matrix {
//     MAT v;
//     Matrix(MAT V) { swap(v, V); }
//     Matrix(int a, int b) { v.resize(a, vector<int>(b, 0)); }
//     Matrix operator*(const Matrix& o) {
//         // assert(v[0].size() == o.v.size());
//         Matrix M(v.size(), o.v[0].size());
//         for (int i = 0; i < v.size(); i++)
//             for (int j = 0; j < o.v[0].size(); j++)
//                 for (int k = 0; k < (int)v[0].size(); k++)
//                     M.v[i][j] += v[i][k] * o.v[k][j];
//         return M;
//     }
//     Matrix operator^(int k) {
//         if (k&1) return (((*this)*(*this))^(k/2))*(*this);
//         else return ((*this)*(*this))^(k/2);
//     }
// };

/*
 * Segment Tree
 * - RANGE QUERY
 * - RANGE UPDATE
*/

// template<class T>
// T identity; // [SET IDENTITY OF CLASS T, hardcode T classname]
// struct ND {
//     ND* ch[2] = { nullptr, nullptr };
//     T v, f;

//     inline void create() {
//         if (!ch[0]) ch[0] = new ND; // [POTENTIALLY UPDATE VALUES]
//         if (!ch[1]) ch[1] = new ND; // [POTENTIALLY UPDATE VALUES]
//     }

//     inline void merge(int l, int r) {
//         // [INSERT CODE FOR MERGE SEGMENT]
//     }

//     inline void push(int l, int r) {
//         create();
//         // [INSERT CODE FOR PUSHING SEGMENT FLAG]
//     }

//     void upd(int l, int r, int L, int R, T K) {
//         push(l, r);
//         if (R < l || r < L) return;
//         if (L <= l && r <= R) {
//             // [INSERT CODE FOR UPDATE SEGMENT]
//             return;
//         }
//         int m = (l + r) >> 1;
//         ch[0]->upd(l, m, L, R, K);
//         ch[1]->upd(m+1, r, L, R, K);
//         merge(l, r);
//     }

//     T qry(int l, int r, int L, int R) {
//         push(l, r);
//         if (R < l || r < L) return identity;
//         if (L <= l && r <= R) return s;
//         int m = (l + r) >> 1;
//         return merge(ch[0]->qry(l,m,L,R), ch[1]->qry(m+1,r,L,R));
//     }

//     ~ND() { delete ch[0]; delete ch[1]; ch[0] = ch[1] = nullptr; }
// };

/*
 * Convex Hull Trick
 * - ORDERED SLOPES
 * - UNORDERED QUERIES O(N log N), ORDERED QUERIES O(N)
*/

// struct F {
//     int a, b;
//     F() {}
//     F(int A, int B) : a(A), b(B){ if (b<0) a*=-1, b*=-1; }
//     bool operator<(F o) const { return a*o.b < b*o.a; }
//     bool operator<=(F o) const { return a*o.b <= b*o.a; }
// };
// struct L {
//     int m, b;
//     int operator()(int x) { return m*x + b; }
//     F operator^(L o) { return F{b-o.b,o.m-m}; }
// };
// struct CHT {
//     vector<L> h;
//     // deque<L> h; // if using SECOND CODE in QRY FUNCTION
//     void add(L l) {
//         // // min hull + decreasing slopes OR max hull + increasing slopes
//         // while (h.size() >= 2 && (h.end()[-2]^l) <= (h.end()[-2]^h.back())) h.pop_back();
//         // h.push_back(l);

//         // // min hull + increasing slopes OR max hull + decreasing slopes
//         // while (h.size() >= 2 && (h.end()[-2]^h.back()) <= (h.end()[-2]^l)) h.pop_back();
//         // h.push_back(l);
//     }
//     int qry(int x) {
//         // O(N log N) time, unordered queries
//         // int lo = 0, hi = h.size()-1;
//         // while (lo < hi) {
//         //     int m = (lo + hi) >> 1;
//         //     if (h[m](x) < h[m+1](x)) // < or > depending on min/max qry
//         //         lo = m+1;
//         //     else 
//         //         hi = m;
//         // }
//         // return h[lo](x);

//         // if need O(N) time, use **DEQUE INSTEAD OF VECTOR** & use code below
//         // while (h.size() >= 2 && h[1](x) < h[0](x)) h.pop_front(); // < or > depending on min/max qry
//         // return h[0](x);
//     }
// };

/*
 * Li Chao Tree 
 * UNORDERED SLOPES, UNORDERED QUERIES O(N log N)
 * Extension of segment tree
*/

// struct F {
//     int a, b;
//     F() {}
//     F(int A, int B) : a(A), b(B){ if (b<0) a*=-1, b*=-1; }
//     bool operator<(F o) const { return a*o.b < b*o.a; }
//     bool operator<=(F o) const { return a*o.b <= b*o.a; }
// };
// struct L {
//     int m, b;
//     int operator()(int x) { return m*x + b; }
//     F operator^(L o) { return F{b-o.b,o.m-m}; }
// };
// // #define OP(a, b) (a < b) // DEFINE OPERATOR
// // #define OPR(a, b) (min(a,b)) // DEFINE OPERATOR
// #define OP(a, b) (a > b)
// #define OPR(a, b) (max(a,b))
// struct ND {
//     ND * ch[2] = { nullptr, nullptr };
//     L v = L{1, 0};
//     void create() {
//         if (!ch[0]) { ch[0] = new ND; ch[0]->v = v; }
//         if (!ch[1]) { ch[1] = new ND; ch[1]->v = v; }
//     }
//     void upd(int l, int r, L V) {
//         // update the li chao tree with line V
//         if (v.m == V.m && v.b == V.b) return;
//         if (l == r) {
//             // if new line is better, swap
//             if (OP(V(l), v(l))) swap(V, v);
//             return;
//         }
//         create();
//         int m = (l + r) >> 1; // midpoint
//         if (OP(V(m), v(m))) swap(V, v); // if new line is better at mid, it has one half, so swap
//         F x = v ^ V; // intersection of two lines
//         if (x <= F(m, 1)) 
//             ch[0]->upd(l, m, V);
//         else
//             ch[1]->upd(m+1, r, V);
//     }
//     int qry(int l, int r, int X) {
//         // qry li chao tree for x coord X        
//         int ret = v(X); // gets value for current segment
//         if (l != r) {
//             int m = (l + r) >> 1;
//             if (X <= m && ch[0] != nullptr) return OPR(ret, ch[0]->qry(l, m, X));
//             if (X > m && ch[1] != nullptr) return OPR(ret, ch[1]->qry(m+1, r, X));
//         }
//         return ret;
//     }
// };
// #undef OP
// #undef OPR

/*
 * Heavy Light Decomp
*/
#define valid(a, b) (a >= 1 && b >= 1 && a <= r && b <= s)

const int maxr = 2005;

int r, s;
int gr[maxr][maxr];
int dp[maxr][maxr], f[maxr][maxr], g[maxr][maxr];

int dr[] = {1, 0, -1, 0};
int dc[] = {0, 1, 0, -1};

pair<int,int> curc;
int num;


void ff(int R, int C) {
    num += gr[R][C]; gr[R][C] = -1;
    curc.first = min(curc.first, C), curc.second = max(curc.second, C);

    for (int i = 0; i < 4; i++) {
        int nr = R + dr[i], nc = C + dc[i];
        if (!valid(nr, nc)) continue;
        if (gr[nr][nc] == -1) continue;
        ff(nr, nc);
    }
}

void dvc(int L, int R, int lh, int hh, int k) {
    if (L > R) return;
    // cout << "-----" << endl;
    int m = (L + R) >> 1;

    // for (int i = lh; i <= min(m, hh); i++)
        // cout << L << ' ' << R << ' ' << m << ' ' << i << ' ' << dp[i][k-1] + f[i][m] << endl;

    int j = lh;
    for (int i = lh; i <= min(m, hh); i++)
        if (dp[m][k] <= dp[i][k-1] + f[i][m])
            dp[m][k] = dp[i][k-1] + f[i][m], j = i;

    if (L == R) return;

    dvc(L, m-1, lh, j, k);
    dvc(m+1, R, j, hh, k);
}

void solve() {
    cin >> r >> s;

    for (int i = 1; i <= r; i++)
        for (int j = 1; j <= s; j++) {
            char c; cin >> c;
            if (c == '.') gr[i][j] = -1;
            else gr[i][j] = c - '0';
        }

    for (int i = 1; i <= r; i++)
        for (int j = 1; j <= s; j++) {
            if (gr[i][j] == -1) continue;
            curc = {j, j}; num = 0;
            ff(i, j);
            for (int k = curc.first; k <= curc.second; k++)
                g[curc.first][k] += num;
        }

    for (int j = s; j >= 1; j--) // loop over js
        for (int i = j-1; i >= 0; i--) // loop over is stemming away from cur j
            f[i][j] = f[i+1][j] + g[i+1][j];

    // for (int i = 0; i <= s; i++) {
    //     for (int j = 0; j <= s; j++) {
    //         cout << g[i][j] << ' ';
    //     }
    //     cout << endl;
    // }

    // cout << endl;

    // for (int i = 0; i <= s; i++) {
    //     for (int j = 0; j <= s; j++) {
    //         cout << f[i][j] << ' ';
    //     }
    //     cout << endl;
    // }

    // cout << endl;

    for (int k = 1; k <= s; k++)
        dvc(1, s, 0, s, k);

    for (int i = 1; i <= s; i++) {
        int r = 0;
        for (int j = 1; j <= s; j++)
            r = max(dp[j][i], r);
        cout << r << endl;
    }


    return;
}

signed main() {
    cin.tie(nullptr) -> ios::sync_with_stdio(false);
    // freopen("main.in", "r", stdin);
    int t; 
    // cin >> t;
    t=1;
    while(t--) solve();
    return 0;
}
# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 1108 KB Output is correct
2 Correct 1 ms 1108 KB Output is correct
3 Correct 1 ms 980 KB Output is correct
4 Correct 1 ms 1108 KB Output is correct
5 Correct 1 ms 1108 KB Output is correct
6 Correct 1 ms 980 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 1108 KB Output is correct
2 Correct 1 ms 1108 KB Output is correct
3 Correct 1 ms 980 KB Output is correct
4 Correct 1 ms 1108 KB Output is correct
5 Correct 1 ms 1108 KB Output is correct
6 Correct 1 ms 980 KB Output is correct
7 Correct 6 ms 6100 KB Output is correct
8 Correct 7 ms 6076 KB Output is correct
9 Correct 7 ms 7508 KB Output is correct
10 Correct 6 ms 6100 KB Output is correct
11 Correct 5 ms 5972 KB Output is correct
12 Correct 5 ms 5460 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 1108 KB Output is correct
2 Correct 1 ms 1108 KB Output is correct
3 Correct 1 ms 980 KB Output is correct
4 Correct 1 ms 1108 KB Output is correct
5 Correct 1 ms 1108 KB Output is correct
6 Correct 1 ms 980 KB Output is correct
7 Correct 6 ms 6100 KB Output is correct
8 Correct 7 ms 6076 KB Output is correct
9 Correct 7 ms 7508 KB Output is correct
10 Correct 6 ms 6100 KB Output is correct
11 Correct 5 ms 5972 KB Output is correct
12 Correct 5 ms 5460 KB Output is correct
13 Correct 270 ms 57496 KB Output is correct
14 Correct 286 ms 57548 KB Output is correct
15 Correct 334 ms 123656 KB Output is correct
16 Correct 247 ms 57448 KB Output is correct
17 Correct 238 ms 54968 KB Output is correct
18 Correct 245 ms 53316 KB Output is correct