Submission #625695

#	Time	Username	Problem	Language	Result	Execution time	Memory
625695		I_love_Hoang_Yen	Catfish Farm (IOI22_fish)	C++17	70 / 100	729 ms	2097152 KiB

fish

#include "bits/stdc++.h"
using namespace std;

#define int long long
#define i_1 jakcjacjl
struct Fish {
    int col, row;
    int weight;
};
bool operator < (const Fish& a, const Fish& b) {
    if (a.col != b.col) return a.col < b.col;
    return a.row < b.row;
}

void upMax(int& f, int val) {
    if (val > f) f = val;
}

// sub1 - 3 {{{
// fishes are on even columns -> build piers on odd columns
// & catch all fishes
int sub1(const std::vector<Fish>& fishes) {
    int res = 0;
    for (const auto& fish : fishes) {
        res += fish.weight;
    }
    return res;
}

// fishes are on first 2 columns
int sub2(int n, const std::vector<Fish>& fishes) {
    std::vector<int> zeroes(n);  // prefix sum of fish weights at column == 0
    std::vector<int> ones(n);    // prefix sum of fish weights at column == 1
    for (const auto& fish : fishes) {
        if (fish.col == 0) zeroes[fish.row] += fish.weight;
        if (fish.col == 1) ones[fish.row] += fish.weight;
    }

    std::partial_sum(zeroes.begin(), zeroes.end(), zeroes.begin());
    std::partial_sum(ones.begin(), ones.end(), ones.begin());

    int res = ones.back();  // init: only catch fishes at column == 1
    for (int i = 0; i < n; ++i) {
        // build pier until at column 1, row 0-i
        if (n == 2) upMax(res, zeroes[i]);
        else upMax(res, zeroes[i] + ones.back() - ones[i]);
    }
    return res;
}

// all fishes are on row == 0
int sub3(int n, const std::vector<Fish>& fishes) {
    std::vector<int> weights(n);  // weights[i] = weight of fish at column i
    for (const auto& fish : fishes) {
        weights[fish.col] += fish.weight;
    }

    // f[i] = best strategy if we BUILD PIER AT i, only considering col 0..i
    // i-4 i-3 i-2 i-1 i
    std::vector<int> f(n);
    f[0] = 0;
    for (int i = 1; i < n; ++i) {
        f[i] = std::max(f[i-1], weights[i-1]);
        if (i >= 2) {
            upMax(f[i], f[i-2] + weights[i-1]);
        }
        if (i >= 3) {
            upMax(f[i], f[i-3] + weights[i-2] + weights[i-1]);
        }
    }

    int res = 0;
    for (int i = 0; i < n; ++i) {
        int cur = f[i];
        if (i + 1 < n) cur += weights[i+1];
        upMax(res, cur);
    }
    return res;
}
// }}}

// sub 5 N <= 300 {{{
int sub5(int n, const std::vector<Fish>& fishes) {
    // Init weights[i][j] = sum of fish on column i, from row 0 -> row j
    std::vector<std::vector<int>> weights(n, std::vector<int> (n, 0));
    for (const auto& fish : fishes) {
        weights[fish.col][fish.row] += fish.weight;
    }
    for (int col = 0; col < n; ++col) {
        std::partial_sum(weights[col].begin(), weights[col].end(), weights[col].begin());
    }

    // f[c][r] = best strategy if we last BUILD PIER AT column c, row r
    //           only considering fishes <= (c, r)
    // g[c][r] = similar to f[c][r] but consider fishes at column c, in row [r, n-1]
    std::vector<std::vector<int>> f(n, std::vector<int> (n, 0)),
                                  g(n, std::vector<int> (n, 0));
    // f <= g
    for (int c = 1; c < n; ++c) {
        for (int r = 0; r < n; ++r) {
            // this is first pier
            f[c][r] = g[c][r] = weights[c-1][r];

            // last pier at column i-1
            for (int lastRow = 0; lastRow < n; ++lastRow) {
                if (lastRow <= r) {
                    int cur = std::max(
                            f[c-1][lastRow] + weights[c-1][r] - weights[c-1][lastRow],
                            g[c-1][lastRow]);
                    upMax(f[c][r], cur);
                    upMax(g[c][r], cur);
                } else {
                    upMax(f[c][r], g[c-1][lastRow]);
                    upMax(g[c][r], g[c-1][lastRow] + weights[c][lastRow] - weights[c][r]);
                }
            }

            // last pier at column i-2
            if (c >= 2) {
                for (int lastRow = 0; lastRow < n; ++lastRow) {
                    int cur = g[c-2][lastRow] + weights[c-1][std::max(lastRow, r)];
                    upMax(f[c][r], cur);
                    upMax(g[c][r], cur);
                }
            }
            
            // last pier at column i-3
            if (c >= 3) {
                for (int lastRow = 0; lastRow < n; ++lastRow) {
                    int cur = g[c-3][lastRow] + weights[c-2][lastRow] + weights[c-1][r];
                    upMax(f[c][r], cur);
                    upMax(g[c][r], cur);
                }
            }
        }
    }

    int res = 0;
    for (int c = 0; c < n; ++c) {
        for (int r = 0; r < n; ++r) {
            assert(g[c][r] >= f[c][r]);
            int cur = g[c][r];
            if (c + 1 < n) {
                cur += weights[c+1][r];
            }
            upMax(res, cur);
        }
    }
    return res;
}
// }}}

// N <= 3000 {{{
int sub6(int n, const std::vector<Fish>& fishes) {
    // Init weights[i][j] = sum of fish on column i, from row 0 -> row j
    std::vector<std::vector<int>> weights(n, std::vector<int> (n, 0));
    for (const auto& fish : fishes) {
        weights[fish.col][fish.row] += fish.weight;
    }
    for (int col = 0; col < n; ++col) {
        std::partial_sum(weights[col].begin(), weights[col].end(), weights[col].begin());
    }

    // f[c][r] = best strategy if we last BUILD PIER AT column c, row r
    //           only considering fishes <= (c, r)
    // g[c][r] = similar to f[c][r] but consider fishes at column c, in row [r, n-1]
    std::vector<std::vector<int>> f(n, std::vector<int> (n, 0)),
                                  g(n, std::vector<int> (n, 0)),
                                  g_prefix_max(n, std::vector<int> (n, 0)),
                                  g_suffix_max(n, std::vector<int> (n, 0)),
                                  g_with_next_col_suffix_max(n, std::vector<int> (n, 0)),
                                  g_with_next_col_prefix_max(n, std::vector<int> (n, 0)),
                                  f_with_next_col_prefix_max(n, std::vector<int> (n, 0));

    // f <= g
    for (int c = 0; c < n; ++c) {
        // compute {{{
        if (c > 0) {
            for (int r = 0; r < n; ++r) {
                // this is first pier
                f[c][r] = g[c][r] = weights[c-1][r];

                // last pier at column i-1
                if (c >= 1) {
                    // last row <= r
                    int cur = std::max(
                            g_prefix_max[c-1][r],
                            f_with_next_col_prefix_max[c-1][r] + weights[c-1][r]);
                    upMax(f[c][r], cur);
                    upMax(g[c][r], cur);

                    // last row > r
                    if (r + 1 < n) {
                        upMax(f[c][r], g_suffix_max[c-1][r + 1]);
                        upMax(g[c][r], g_with_next_col_suffix_max[c-1][r + 1] - weights[c][r]);
                    }
                }

                // last pier at column i-2
                if (c >= 2) {
                    int cur = std::max(
                            g_prefix_max[c-2].back() + weights[c-1][r],
                            g_with_next_col_prefix_max[c-2].back());
                    upMax(f[c][r], cur);
                    upMax(g[c][r], cur);
                }

                // last pier at column i-3
                if (c >= 3) {
                    int cur = g_with_next_col_prefix_max[c-3].back() + weights[c-1][r];
                    upMax(f[c][r], cur);
                    upMax(g[c][r], cur);
                }
            }
        }
        // }}}
        
        // aggregate {{{
        // g_prefix_max[c][r] = max(g[c][0], .., g[c][r])
        auto MAX = [] (auto a, auto b) { return std::max(a, b); };
        std::partial_sum(g[c].begin(), g[c].end(), g_prefix_max[c].begin(), MAX);
        std::partial_sum(g[c].rbegin(), g[c].rend(), g_suffix_max[c].rbegin(), MAX);

        if (c + 1 < n) {
            // g_with_next_col_prefix_max[c][r] = max(
            //     g[c][0] + weights[c+1][0],
            //     ...
            //     g[c][r] + weights[c+1][r])
            for (int r = 0; r < n; ++r) {
                g_with_next_col_prefix_max[c][r] = g[c][r] + weights[c+1][r];
            }
            g_with_next_col_suffix_max[c] = g_with_next_col_prefix_max[c];

            std::partial_sum(
                    g_with_next_col_prefix_max[c].begin(),
                    g_with_next_col_prefix_max[c].end(),
                    g_with_next_col_prefix_max[c].begin(),
                    MAX);
            std::partial_sum(
                    g_with_next_col_suffix_max[c].rbegin(),
                    g_with_next_col_suffix_max[c].rend(),
                    g_with_next_col_suffix_max[c].rbegin(),
                    MAX);

            for (int r = 0; r < n; ++r) {
                f_with_next_col_prefix_max[c][r] = f[c][r] - weights[c][r];
            }
            std::partial_sum(
                    f_with_next_col_prefix_max[c].begin(),
                    f_with_next_col_prefix_max[c].end(),
                    f_with_next_col_prefix_max[c].begin(),
                    MAX);
        }
        // }}}
    }

    int res = 0;
    for (int c = 0; c < n; ++c) {
        for (int r = 0; r < n; ++r) {
            assert(g[c][r] >= f[c][r]);
            int cur = g[c][r];
            if (c + 1 < n) {
                cur += weights[c+1][r];
            }
            upMax(res, cur);
        }
    }
    return res;

}
// }}}

/*
int sub7(int n, const std::vector<Fish>& fishes) {
    // fishesAt[col] = vector storing all fishes at column `col`
    std::vector<std::vector<std::pair<int, int>>> fishesAt(n);
    std::vector<std::vector<int>> weights(n);

    for (const auto& fish : fishes) {
        fishesAt[fish.col].push_back({fish.row, fish.weight});
    }

    for (int c = 0; c < n; ++c) {
        fishesAt[c].push_back({-1, 0});
        std::sort(fishesAt[c].begin(), fishesAt[c].end());
        fishesAt[c].push_back({1000111, 0});

        for (auto [row, weight] : fishesAt[c]) {
            weights[c].push_back(weight);
        }
        std::partial_sum(weights[c].begin(), weights[c].end(), weights[c].begin());
    }

    // f[c][r] = best strategy if we last BUILD PIER AT column c, row r
    //           only considering fishes <= (c, r)
    // g[c][r] = similar to f[c][r] but consider fishes at column c, in row [r, n-1]
    std::vector<std::vector<int>> f(n),
                                  g(n),
                                  g_prefix_max(n),
                                  g_suffix_max(n),
                                  g_with_next_col_suffix_max(n),
                                  g_with_next_col_prefix_max(n),
                                  f_with_next_col_prefix_max(n);

    for (int c = 0; c < n; ++c) {
        // compute {{{
        int k = (int) fishesAt[c].size();
        f[c].resize(k);
        g[c].resize(k);

        if (c > 0) {
            for (int i = 0; i < k - 1; ++i) {  // ignore last
                int r = fishesAt[c][i].first;
                int i_1 = lower_bound(
                        fishesAt[c-1].begin(), fishesAt[c-1].end(), std::make_pair(r+1, -1LL))
                    - fishesAt[c-1].begin();
                --i_1;
                assert(fishesAt[c-1][i_1].first <= r && fishesAt[c-1][i_1 + 1].first > r);

#define r ajckajcl

                // this is first pier
                f[c][i] = g[c][i] = weights[c-1][i_1];

                // last pier at column i-1
                {
                    // last row <= r
                    int cur = std::max(
                            g_prefix_max[c-1][i_1],
                            f_with_next_col_prefix_max[c-1][i_1] + weights[c-1][i_1]);
                    f[c][i] = std::max(f[c][i], cur);
                    g[c][i] = std::max(g[c][i], cur);

                    // last row > r
                    if (r + 1 < k-1) {
                        f[c][i] = std::max(f[c][i], 
                    }
                }

#undef r
            }
        }
        // }}}
    }
    return 0;
}
*/

#undef int
long long max_weights(
        int n, int nFish,
        std::vector<int> xs,
        std::vector<int> ys,
        std::vector<int> ws) {
    std::vector<Fish> fishes;
    for (int i = 0; i < nFish; ++i) {
        fishes.push_back({xs[i], ys[i], ws[i]});
    }

    if (std::all_of(xs.begin(), xs.end(), [] (int x) { return x % 2 == 0; })) {
        return sub1(fishes);
    }
    if (*std::max_element(xs.begin(), xs.end()) <= 1) {
        return sub2(n, fishes);
    }
    if (*std::max_element(ys.begin(), ys.end()) == 0) {
        return sub3(n, fishes);
    }
    if (n <= 300) return sub5(n, fishes);
    return sub6(n, fishes);
}

• Subtask 1: 3 / 3
• Subtask 2: 6 / 6
• Subtask 3: 9 / 9
• Subtask 4: 14 / 14
• Subtask 5: 21 / 21
• Subtask 6: 17 / 17
• Subtask 7: 0 / 14
• Subtask 8: 0 / 16