Submission #1230642

#	Time	Username	Problem	Language	Result	Execution time	Memory
1230642		brito.joao	Colouring a rectangle (eJOI19_colouring)	C++20	60 / 100	198 ms	167936 KiB

colouring

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>

using namespace std;

const long long INF = 1e18;

// Structure to represent a directed edge in the flow network
struct Edge {
    int to;
    long long capacity;
    int rev; // Index of the reverse edge in the adjacency list of 'to'
};

vector<vector<Edge>> adj;
vector<int> level;
vector<size_t> iter;
vector<pair<int, int>> seg_tree_children; // Stores children for segment tree nodes
vector<int> v_nodes; // Stores the indices of v-nodes for one parity

// Adds a directed edge and its residual edge
void add_edge(int u, int v, long long cap) {
    adj[u].push_back({v, cap, (int)adj[v].size()});
    adj[v].push_back({u, 0, (int)adj[u].size() - 1});
}

// BFS to build the level graph for Dinic's algorithm
bool bfs(int s, int t) {
    level.assign(adj.size(), -1);
    vector<int> q;
    q.push_back(s);
    level[s] = 0;
    size_t head = 0;
    while(head < q.size()){
        int u = q[head++];
        for(const auto& edge : adj[u]){
            if(edge.capacity > 0 && level[edge.to] < 0){
                level[edge.to] = level[u] + 1;
                q.push_back(edge.to);
            }
        }
    }
    return level[t] != -1;
}

// DFS to find augmenting paths in the level graph
long long dfs(int u, int t, long long f) {
    if (u == t) return f;
    for (size_t& i = iter[u]; i < adj[u].size(); ++i) {
        Edge& e = adj[u][i];
        if (e.capacity > 0 && level[u] < level[e.to]) {
            long long d = dfs(e.to, t, min(f, e.capacity));
            if (d > 0) {
                e.capacity -= d;
                adj[e.to][e.rev].capacity += d;
                return d;
            }
        }
    }
    return 0;
}

// Dinic's max-flow algorithm
long long max_flow(int s, int t) {
    long long flow = 0;
    while (bfs(s, t)) {
        iter.assign(adj.size(), 0);
        long long f;
        while ((f = dfs(s, t, INF)) > 0) {
            flow += f;
        }
    }
    return flow;
}


int build_seg_tree(int l, int r, int& node_idx) {
    int curr_node = node_idx++;
    seg_tree_children.emplace_back(-1, -1);
    if (l == r) {
        add_edge(curr_node, v_nodes[l], INF);
    } else {
        int mid = l + (r - l) / 2;
        int child1 = build_seg_tree(l, mid, node_idx);
        int child2 = build_seg_tree(mid + 1, r, node_idx);
        seg_tree_children[curr_node - (int)v_nodes.back() - 1] = {child1, child2};
        add_edge(curr_node, child1, INF);
        add_edge(curr_node, child2, INF);
    }
    return curr_node;
}

void query_seg_tree(int node_idx, int seg_base_idx, int l, int r, int ql, int qr, int u_node) {
     if (ql > r || qr < l) {
        return;
    }
    if (ql <= l && r <= qr) {
        add_edge(u_node, node_idx, INF);
        return;
    }
    int mid = l + (r - l) / 2;
    pair<int,int> children = seg_tree_children[node_idx - seg_base_idx];
    if(children.first != -1) {
        query_seg_tree(children.first, seg_base_idx, l, mid, ql, qr, u_node);
        query_seg_tree(children.second, seg_base_idx, mid + 1, r, ql, qr, u_node);
    }
}


long long solve_for_parity(int m, int n, const vector<long long>& c1, const vector<long long>& c2, int parity) {
    vector<long long> costs1, costs2;
    vector<int> i_orig, j_orig;
    
    for (int i = 1 - n; i <= m - 1; ++i) {
        if (abs(i % 2) == parity) {
            i_orig.push_back(i);
            costs1.push_back(c1[i + n - 1]);
        }
    }

    for (int j = 0; j <= m + n - 2; ++j) {
        if (j % 2 == parity) {
            j_orig.push_back(j);
            costs2.push_back(c2[j]);
        }
    }
    
    if (costs1.empty() || costs2.empty()) return 0;

    int num_u = costs1.size();
    int num_v = costs2.size();
    int num_seg_nodes_est = 4 * num_v;
    int total_nodes = 2 + num_u + num_v + num_seg_nodes_est;

    adj.assign(total_nodes, vector<Edge>());
    int S = 0, T = 1;
    vector<int> u_nodes(num_u);
    v_nodes.assign(num_v, 0);

    int node_idx = 2;
    for (int i = 0; i < num_u; ++i) u_nodes[i] = node_idx++;
    for (int i = 0; i < num_v; ++i) v_nodes[i] = node_idx++;
    
    for (int i = 0; i < num_u; ++i) add_edge(S, u_nodes[i], costs1[i]);
    for (int i = 0; i < num_v; ++i) add_edge(v_nodes[i], T, costs2[i]);

    int seg_base_idx = node_idx;
    seg_tree_children.clear();
    if(num_v > 0) seg_tree_children.reserve(num_seg_nodes_est);
    
    int seg_root = -1;
    if (num_v > 0) {
        seg_root = build_seg_tree(0, num_v - 1, node_idx);
    }
    
    if(num_v > 0) seg_tree_children.resize(node_idx - seg_base_idx);

    for (int u_idx = 0; u_idx < num_u; ++u_idx) {
        int i_val = i_orig[u_idx];
        
        long long j_start_val = max((long long)i_val, (long long)-i_val);
        long long j_end_val = min(2LL * m - 1 - i_val, 2LL * n - 1 + i_val);

        if (j_start_val > j_end_val) continue;
        
        auto it_start = lower_bound(j_orig.begin(), j_orig.end(), j_start_val);
        if (it_start == j_orig.end() || *it_start > j_end_val) continue;
        int v_idx_start = distance(j_orig.begin(), it_start);

        auto it_end = upper_bound(j_orig.begin(), j_orig.end(), j_end_val);
        --it_end;
        int v_idx_end = distance(j_orig.begin(), it_end);
        
        if(v_idx_start <= v_idx_end && seg_root != -1) {
            query_seg_tree(seg_root, seg_base_idx, 0, num_v - 1, v_idx_start, v_idx_end, u_nodes[u_idx]);
        }
    }
    adj.resize(node_idx);
    return max_flow(S, T);
}

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

    int m, n;
    cin >> m >> n;
    vector<long long> c1(m + n - 1), c2(m + n - 1);
    for (int i = 0; i < m + n - 1; ++i) cin >> c1[i];
    for (int i = 0; i < m + n - 1; ++i) cin >> c2[i];

    long long total_cost = 0;
    total_cost += solve_for_parity(m, n, c1, c2, 0); // Even parity
    total_cost += solve_for_parity(m, n, c1, c2, 1); // Odd parity
    
    cout << total_cost << endl;

    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 20 / 20
• Subtask 5: 10 / 10
• Subtask 6: 0 / 20
• Subtask 7: 0 / 20