oj.uz

답안 #130832

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
130832 2019-07-16T06:48:03 Z xantho Election Campaign (JOI15_election_campaign) C++17
100 / 100
 330 ms 31528 KB 
#include <algorithm>
#include <iostream>
#include <vector>

class heavy_light {
    struct path {
        int u, v;
        int lca;
        long long value;
    };

    enum class chain_value_type {
        children,
        vertex
    };

    const int UNCHECKED = -1;
    const int ROOT = 0; // Standard root

    int num_vertices;
    std::vector<std::vector<int>> adj_list;
    std::vector<int> parent; // Parent of each vertex in rooted tree.
    std::vector<int> depth; // Depth of each vertex.
    std::vector<int> dfs_start; // Time when a vertex's DFS processing starts. Used to check for Ancestors
    std::vector<int> dfs_end; // Time when a vertex's DFS processing ends. Used to check for Ancestors.
    std::vector<int> subtree_size; // Number of vertices in subtree rooted at each vertex.
    std::vector<std::vector<int>> chains; // The i-th vector is the i-th chain
    std::vector<int> chain_index; // Which chain is a vertex part of? (Starts from 0)
    std::vector<int> position_in_chain; // How deep is a vertex in the chain? (0 is closest to the root)

    // DP Variables
    std::vector<std::vector<path>> lca_paths; // The i-th vector represents all paths that has i as the LCA
    std::vector<std::vector<long long>> cumulative_chain_children_memo; // The j-th index of the i-th vector represents the sum of the children's DP values for the j-th vertex and beyond within the i-th chain
    std::vector<std::vector<long long>> cumulative_chain_vertex_memo; // Exactly the same as above, but stores the vertex's DP value instead of the children's

    void dfs(int u, int& time) {
        dfs_start[u] = time;
        subtree_size[u] = 1;

        for (int v : adj_list[u]) {
            if (v == parent[u]) {
                continue;
            }

            parent[v] = u;
            depth[v] = depth[u] + 1;

            ++time;
            dfs(v, time);

            subtree_size[u] += subtree_size[v];
        }

        dfs_end[u] = time;
        ++time;
    }

    void make_rooted_tree() {
        parent[ROOT] = ROOT;
        depth[ROOT] = 0;

        int time = 0;
        dfs(ROOT, time);
    }

    void start_new_chain(int head) {
        int new_chain_id = chains.size();

        chains.push_back({head});
        chain_index[head] = new_chain_id;
        position_in_chain[head] = 0;
    }

    void recursive_construct_chain(int u) {
        int next_in_chain = UNCHECKED;
        for (int v : adj_list[u]) {
            if (v == parent[u]) {
                continue;
            }

            if (next_in_chain == UNCHECKED || subtree_size[next_in_chain] < subtree_size[v]) {
                next_in_chain = v;
            }
        }

        // Case: No children
        if (next_in_chain == UNCHECKED) {
            return;
        }

        // Set up next vertex in chain
        chains[chain_index[u]].push_back(next_in_chain);
        chain_index[next_in_chain] = chain_index[u];
        position_in_chain[next_in_chain] = position_in_chain[u] + 1;
        recursive_construct_chain(next_in_chain);

        // Start new chains in all other children
        for (int v : adj_list[u]) {
            if (v == parent[u] || v == next_in_chain) {
                continue;
            }

            start_new_chain(v);
            recursive_construct_chain(v);
        }
    }

    void construct_chains() {
        start_new_chain(ROOT);
        recursive_construct_chain(ROOT);
    }

    // Returns true if u is an ancestor of v
    bool is_ancestor(int u, int v) {
        return dfs_start[u] <= dfs_start[v] && dfs_end[v] <= dfs_end[u];
    }

    int find_lca(int u, int v) {
        // We don't really need HLD to find the LCA. 
        // This can be done in O(1) per query using a Sparse Table.
        // (But I didn't know at the time, so oh well)
        
        // Step 1: Find chain containing LCA
        int u_chain_index = chain_index[u];
        int u_curr = chains[u_chain_index][0];

        while (!is_ancestor(u_curr, v)) {
            u_curr = parent[u_curr]; // Move to parent chain
            u_chain_index = chain_index[u_curr];
            u_curr = chains[u_chain_index][0]; // Move to head of chain
        }

        int lca_chain_index = chain_index[u_curr];

        int low = 0;
        int high = (int) (chains[lca_chain_index].size()) - 1;
        int lca = u_curr;

        // Step 2: Perform Binary Search on the chain
        while (low <= high) {
            int mid = low + (high - low) / 2;
            int vertex = chains[lca_chain_index][mid];

            if (is_ancestor(vertex, v) && is_ancestor(vertex, u)) {
                lca = vertex;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }

        return lca;
    }

    long long chain_sum(chain_value_type type, int chain_index, int chain_pos_start, int chain_pos_end) {
        auto &cumulative_memo = (type == chain_value_type::children 
                ? cumulative_chain_children_memo 
                : cumulative_chain_vertex_memo);
        
        auto &chain = cumulative_memo[chain_index];
        return chain[chain_pos_start] - chain[chain_pos_end + 1];
    }

    long long process_path(const path& p) {
        /*
        Here's the idea behind how this works. 
        
        Consider the following tree, and the path from A1 to A5 (i.e. A1-A2-A3-A4-A5):
        
                         A3
                       /    \
                     A2      A4
                    /  \    /  \
                   A1  X3  X4   A5
                  / |          / | \
                 X1 X2        X5 X6 X7
        
        Let DP[V] represent the highest possible score (i.e. number of votes) obtainable
        among all the paths that exist in the subtree rooted at the vertex V.
        
        If we decide to take the path from A1 to A5, then the maximum remaining score we can
        obtain is the sum of the DP values of the remaining untouched children. In this case, 
        that would be
        
        score(A1-A2-A3-A4-A5) + DP[X1] + DP[X2] + DP[X3] + DP[X4] + DP[X5] + DP[X6] + DP[X7]
        
        The DP terms can simply be added together because all the subtrees rooted at those 
        vertices are disjoint.
        
        This is still hard to deal with, since there may be many children to consider, so 
        let's try to represent it in a different way.
        
        Let children(V) represent the set of children of a vertex. For example, children(A2)
        returns {A1, X3}. Also for simplicity, let DP[{U1, ... , Uk}] = DP[U1]  + ... + DP[Uk].
        
        Now, we can represent the expression as such:
        
        score(A1-A2-A3-A4-A5) 
        + DP[children(A1)] + DP[children(A2)] + DP[children(A3)] + DP[children(A4)] + DP[children(A5)]
        - DP[A1] - DP[A2] - DP[A4] - DP[A5]
        
        Or, to make the expression more general, for a path P = A-...-L-...-B, where L is the 
        LCA of vertices A and B:
        
        score(P) + sum(DP[children(v)] for v in P) - sum(DP[v] for v in P excluding L)
        */

        int lca_chain_index = chain_index[p.lca];
        int lca_pos_in_chain = position_in_chain[p.lca];

        long long answer = 0;

        int endpoints[] = {p.u, p.v};
        for (int vertex : endpoints) {
            // Step 1: Climb up from curr to LCA
            long long path_value = 0;

            // Step 1a: Climb until the same chain as LCA
            int curr = vertex;
            while (chain_index[curr] != lca_chain_index) {
                int curr_chain_index = chain_index[curr];
                int curr_pos_in_chain = position_in_chain[curr];
                int curr_chain_head = chains[curr_chain_index][0];

                path_value += chain_sum(chain_value_type::children, curr_chain_index, 0, curr_pos_in_chain);
                // cumulative_chain_children_memo[curr_chain_index][0] - cumulative_chain_children_memo[curr_chain_index][curr_pos_in_chain + 1];
                path_value -= chain_sum(chain_value_type::vertex, curr_chain_index, 0, curr_pos_in_chain);
                // path_value -= cumulative_chain_vertex_memo[curr_chain_index][0] - cumulative_chain_vertex_memo[curr_chain_index][curr_pos_in_chain + 1];

                curr = parent[curr_chain_head];
            }

            // Step 1b: Climb from curr to LCA
            {
                int curr_pos_in_chain = position_in_chain[curr];
                path_value += chain_sum(chain_value_type::children, lca_chain_index, lca_pos_in_chain + 1, curr_pos_in_chain);
                path_value -= chain_sum(chain_value_type::vertex, lca_chain_index, lca_pos_in_chain + 1, curr_pos_in_chain);
                // path_value += cumulative_chain_children_memo[lca_chain_index][lca_pos_in_chain + 1] - cumulative_chain_children_memo[lca_chain_index][curr_pos_in_chain + 1];
                // path_value -= cumulative_chain_vertex_memo[lca_chain_index][lca_pos_in_chain + 1] - cumulative_chain_vertex_memo[lca_chain_index][curr_pos_in_chain + 1];
            }

            answer += path_value;
        }

        answer += chain_sum(chain_value_type::children, lca_chain_index, lca_pos_in_chain, lca_pos_in_chain) + p.value;

        // answer += (cumulative_chain_children_memo[lca_chain_index][lca_pos_in_chain] - cumulative_chain_children_memo[lca_chain_index][lca_pos_in_chain + 1]) + p.value;
        return answer;
    }

    long long bottom_up_dp(int u) {
        // DP value of u represents the maximum value obtainable within subtree rooted at u.

        // Step 1: Process all children first. Keep track of sum of DP values of children
        long long children_dp = 0;
        for (int v : adj_list[u]) {
            if (v == parent[u]) {
                continue;
            }

            children_dp += bottom_up_dp(v);
        }

        // Step 2: Update cumulative chain children memo
        int u_chain_index = chain_index[u];
        int u_pos_in_chain = position_in_chain[u];

        cumulative_chain_children_memo[u_chain_index][u_pos_in_chain] = cumulative_chain_children_memo[u_chain_index][u_pos_in_chain + 1] + children_dp;

        // Step 3: Process all paths to find DP value of u
        long long u_dp_value = children_dp;

        for (path& p : lca_paths[u]) {
            u_dp_value = std::max(u_dp_value, process_path(p));
        }

        // Step 4: Update cumulative chain vertex memo
        cumulative_chain_vertex_memo[u_chain_index][u_pos_in_chain] = cumulative_chain_vertex_memo[u_chain_index][u_pos_in_chain + 1] + u_dp_value;

        // Step 5: Return
        return u_dp_value;
    }

public:
    heavy_light(int _num_vertices) :
            num_vertices(_num_vertices),
            adj_list(num_vertices),
            parent(num_vertices, UNCHECKED),
            depth(num_vertices, UNCHECKED),
            dfs_start(num_vertices, UNCHECKED),
            dfs_end(num_vertices, UNCHECKED),
            subtree_size(num_vertices, UNCHECKED),
            chain_index(num_vertices, UNCHECKED),
            position_in_chain(num_vertices, UNCHECKED),
            lca_paths(num_vertices) {}

    void add_edge(int u, int v) {
        adj_list[u].push_back(v);
        adj_list[v].push_back(u);
    }

    void decompose() {
        make_rooted_tree();
        construct_chains();
    }

    void print_details() {
        for (int i = 0; i < (int) chains.size(); ++i) {
            std::cout << "Chain #" << i << ": ";
            for (int u : chains[i]) {
                std::cout << u << "-";
            }
            std::cout << "\n";
        }
        std::cout << "\n";

        for (int i = 0; i < num_vertices; ++i) {
            std::cout << "Vertex #" << i
                    << " - DFS times: " << dfs_start[i] << " to " << dfs_end[i]
                    << ", Chain ID: " << chain_index[i]
                    << ", Position in Chain: " << position_in_chain[i] << "\n";
        }
    }
    void add_path(int u, int v, long long value) {
        int lca = find_lca(u, v);
        lca_paths[lca].push_back({u, v, lca, value});
    }

    long long get_answer() {
        for (auto& chain : chains) {
            cumulative_chain_children_memo.emplace_back(chain.size() + 1, 0);
            cumulative_chain_vertex_memo.emplace_back(chain.size() + 1, 0);
        }

        return bottom_up_dp(ROOT);
    }
};

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int num_vertices;
    std::cin >> num_vertices;

    heavy_light hld(num_vertices);

    for (int i = 0; i < num_vertices - 1; ++i) {
        int u, v;
        std::cin >> u >> v;

        hld.add_edge(u - 1, v - 1); // 1-indexed vertices
    }

    hld.decompose();

    int num_paths;
    std::cin >> num_paths;

    for (int i = 0; i < num_paths; ++i) {
        int u, v;
        long long value;

        std::cin >> u >> v >> value;

        hld.add_path(u - 1, v - 1, value); // 1-indexed vertices
    }

    std::cout << hld.get_answer() << "\n";
}

Subtask #1
10.0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 252 KB Output is correct
3 Correct 2 ms 376 KB Output is correct
4 Correct 3 ms 632 KB Output is correct
5 Correct 131 ms 20140 KB Output is correct
6 Correct 62 ms 21876 KB Output is correct
7 Correct 115 ms 22316 KB Output is correct
8 Correct 121 ms 28880 KB Output is correct
9 Correct 110 ms 20544 KB Output is correct
10 Correct 127 ms 28708 KB Output is correct

Subtask #2
5.0 / 5.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 256 KB Output is correct
2 Correct 2 ms 376 KB Output is correct
3 Correct 3 ms 504 KB Output is correct
4 Correct 136 ms 25300 KB Output is correct
5 Correct 138 ms 25456 KB Output is correct
6 Correct 137 ms 25328 KB Output is correct
7 Correct 136 ms 25328 KB Output is correct
8 Correct 135 ms 25300 KB Output is correct
9 Correct 134 ms 25328 KB Output is correct
10 Correct 152 ms 25328 KB Output is correct

Subtask #3
5.0 / 5.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 256 KB Output is correct
2 Correct 2 ms 376 KB Output is correct
3 Correct 3 ms 504 KB Output is correct
4 Correct 136 ms 25300 KB Output is correct
5 Correct 138 ms 25456 KB Output is correct
6 Correct 137 ms 25328 KB Output is correct
7 Correct 136 ms 25328 KB Output is correct
8 Correct 135 ms 25300 KB Output is correct
9 Correct 134 ms 25328 KB Output is correct
10 Correct 152 ms 25328 KB Output is correct
11 Correct 14 ms 1400 KB Output is correct
12 Correct 144 ms 25328 KB Output is correct
13 Correct 143 ms 25392 KB Output is correct
14 Correct 141 ms 25368 KB Output is correct
15 Correct 143 ms 25340 KB Output is correct
16 Correct 142 ms 25300 KB Output is correct
17 Correct 143 ms 25324 KB Output is correct
18 Correct 138 ms 25328 KB Output is correct
19 Correct 140 ms 25328 KB Output is correct
20 Correct 139 ms 25328 KB Output is correct

Subtask #4
30.0 / 30.0

# 결과 실행 시간 메모리 Grader output
1 Correct 274 ms 22976 KB Output is correct
2 Correct 135 ms 25332 KB Output is correct
3 Correct 218 ms 25292 KB Output is correct
4 Correct 207 ms 31372 KB Output is correct
5 Correct 216 ms 24620 KB Output is correct
6 Correct 207 ms 31528 KB Output is correct
7 Correct 217 ms 24364 KB Output is correct
8 Correct 286 ms 23256 KB Output is correct
9 Correct 154 ms 25328 KB Output is correct
10 Correct 215 ms 23240 KB Output is correct

Subtask #5
10.0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 252 KB Output is correct
3 Correct 2 ms 376 KB Output is correct
4 Correct 3 ms 632 KB Output is correct
5 Correct 131 ms 20140 KB Output is correct
6 Correct 62 ms 21876 KB Output is correct
7 Correct 115 ms 22316 KB Output is correct
8 Correct 121 ms 28880 KB Output is correct
9 Correct 110 ms 20544 KB Output is correct
10 Correct 127 ms 28708 KB Output is correct
11 Correct 3 ms 632 KB Output is correct
12 Correct 3 ms 632 KB Output is correct
13 Correct 3 ms 636 KB Output is correct
14 Correct 3 ms 632 KB Output is correct
15 Correct 3 ms 632 KB Output is correct
16 Correct 3 ms 632 KB Output is correct
17 Correct 3 ms 632 KB Output is correct
18 Correct 3 ms 632 KB Output is correct
19 Correct 3 ms 632 KB Output is correct
20 Correct 3 ms 504 KB Output is correct

Subtask #6
40.0 / 40.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 252 KB Output is correct
3 Correct 2 ms 376 KB Output is correct
4 Correct 3 ms 632 KB Output is correct
5 Correct 131 ms 20140 KB Output is correct
6 Correct 62 ms 21876 KB Output is correct
7 Correct 115 ms 22316 KB Output is correct
8 Correct 121 ms 28880 KB Output is correct
9 Correct 110 ms 20544 KB Output is correct
10 Correct 127 ms 28708 KB Output is correct
11 Correct 2 ms 256 KB Output is correct
12 Correct 2 ms 376 KB Output is correct
13 Correct 3 ms 504 KB Output is correct
14 Correct 136 ms 25300 KB Output is correct
15 Correct 138 ms 25456 KB Output is correct
16 Correct 137 ms 25328 KB Output is correct
17 Correct 136 ms 25328 KB Output is correct
18 Correct 135 ms 25300 KB Output is correct
19 Correct 134 ms 25328 KB Output is correct
20 Correct 152 ms 25328 KB Output is correct
21 Correct 14 ms 1400 KB Output is correct
22 Correct 144 ms 25328 KB Output is correct
23 Correct 143 ms 25392 KB Output is correct
24 Correct 141 ms 25368 KB Output is correct
25 Correct 143 ms 25340 KB Output is correct
26 Correct 142 ms 25300 KB Output is correct
27 Correct 143 ms 25324 KB Output is correct
28 Correct 138 ms 25328 KB Output is correct
29 Correct 140 ms 25328 KB Output is correct
30 Correct 139 ms 25328 KB Output is correct
31 Correct 274 ms 22976 KB Output is correct
32 Correct 135 ms 25332 KB Output is correct
33 Correct 218 ms 25292 KB Output is correct
34 Correct 207 ms 31372 KB Output is correct
35 Correct 216 ms 24620 KB Output is correct
36 Correct 207 ms 31528 KB Output is correct
37 Correct 217 ms 24364 KB Output is correct
38 Correct 286 ms 23256 KB Output is correct
39 Correct 154 ms 25328 KB Output is correct
40 Correct 215 ms 23240 KB Output is correct
41 Correct 3 ms 632 KB Output is correct
42 Correct 3 ms 632 KB Output is correct
43 Correct 3 ms 636 KB Output is correct
44 Correct 3 ms 632 KB Output is correct
45 Correct 3 ms 632 KB Output is correct
46 Correct 3 ms 632 KB Output is correct
47 Correct 3 ms 632 KB Output is correct
48 Correct 3 ms 632 KB Output is correct
49 Correct 3 ms 632 KB Output is correct
50 Correct 3 ms 504 KB Output is correct
51 Correct 270 ms 23328 KB Output is correct
52 Correct 141 ms 25328 KB Output is correct
53 Correct 218 ms 23228 KB Output is correct
54 Correct 198 ms 31400 KB Output is correct
55 Correct 280 ms 23052 KB Output is correct
56 Correct 142 ms 25412 KB Output is correct
57 Correct 216 ms 24024 KB Output is correct
58 Correct 195 ms 31412 KB Output is correct
59 Correct 328 ms 23212 KB Output is correct
60 Correct 154 ms 25328 KB Output is correct
61 Correct 207 ms 24108 KB Output is correct
62 Correct 203 ms 31284 KB Output is correct
63 Correct 330 ms 22828 KB Output is correct
64 Correct 144 ms 25328 KB Output is correct
65 Correct 211 ms 24028 KB Output is correct
66 Correct 205 ms 31396 KB Output is correct
67 Correct 292 ms 22828 KB Output is correct
68 Correct 141 ms 25328 KB Output is correct
69 Correct 222 ms 22836 KB Output is correct
70 Correct 193 ms 31396 KB Output is correct