Submission #607173

#	Time	Username	Problem	Language	Result	Execution time	Memory
607173		jophyyjh	One-Way Streets (CEOI17_oneway)	C++14	30 / 100	14 ms	468 KiB

oneway

/**
 * Notes during contest.
 * 
 * ------ A ------
 * Looks like that it's about strongly-connected components, cycles or bridges. I
 * think we can detect all cycles and merge all those nodes into one. The resulting
 * graph would be a tree, so we can then use a difference array to find each edge's
 * direction. Note that any edge in a cycle is B, and any edge not in cycles is a
 * "bridge".
 *          !IMPORTANT! (Reminded by tinjyu during contest) !IMPORTANT!
 *                          Graph may not be CONNECTED.
 * 
 * ------ B ------
 * We iterate through all cost k. We can take at most k bets. If we can guarantee
 * a profit of p, min(#A_is_the_outcome, #B_is_the_outcome) >= p + k. Binary search
 * is ok.
 * For each k, we wanna find the max achievable for
 * min(#A_is_outcome, #B_is_outcome) - k.
 * 
 * ------ C ------
 * 
 * Time Complexity 1: O(m^2)
 * Time Complexity 2: O(n * log(n * VAL_MAX) * log(n))
 * Time Complexity 3: O()
 * Implementation 0.5       (test if my solution is correct, not optimized, slow)
*/
 
#include <bits/stdc++.h>
 
typedef int64_t     int_t;
typedef std::vector<int_t>  vec;
 
const int_t INF = 0x3f3f3f3f3f3f;
 
// a DSU impl that allows assigning a value to each set
struct DSU {
    vec set, rank;
    int_t components;
    DSU(int_t n) {
        set.resize(n);
        rank.resize(n);
        for (int_t k = 0; k < n; k++)
            set[k] = k, rank[k] = 1;
        components = n;
    }
    inline int_t find(int_t k) {
        if (set[k] == k)
            return k;
        return set[k] = find(set[k]);
    }
    bool merge(int_t a, int_t b) {      // merge b into a   (keep a's value)
        a = find(a), b = find(b);
        if (a == b)
            return false;
        if (rank[b] > rank[a])
            std::swap(a, b);
        set[b] = a, rank[a] += rank[b], components--;
        return true;
    }
};
 
 
struct edge_t {
    int_t a, b;
};
 
struct neighb_t {
    int_t neighb, edge_idx;
};
 
typedef std::vector<neighb_t>   adj_list_t;
 
int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);
 
    int_t n, m;
    std::cin >> n >> m;
    if (n > 3000) {
        std::cout << "bye" << std::endl;
        return 0;
    }
    std::vector<edge_t> original_graph(m);
    for (int_t k = 0; k < m; k++) {
        std::cin >> original_graph[k].a >> original_graph[k].b;
        original_graph[k].a--, original_graph[k].b--;
    }
 
 
    // TODO: improve bridge finding
    std::vector<bool> is_bridge(m);
    // std::cerr << "[debug] is_bridge[]: ";
    for (int_t k = 0; k < m; k++) {
        DSU dsu(n);
        for (int_t i = 0; i < m; i++) {
            if (i == k)
                continue;
            dsu.merge(original_graph[i].a, original_graph[i].b);
        }
        int_t c = dsu.components;
        dsu.merge(original_graph[k].a, original_graph[k].b);
        assert(0 <= c - dsu.components && c - dsu.components <= 1);
        is_bridge[k] = (c != dsu.components);
        // std::cerr << is_bridge[k];
    }
    // std::cerr << std::endl;
 
    DSU compress(n);    // used to compress the nodes
    for (int_t k = 0; k < m; k++) {
        if (!is_bridge[k])
            compress.merge(original_graph[k].a, original_graph[k].b);
    }
    vec map(n, -1);         // mapping an original node into one in our new graph
    // std::cerr << "[debug] map[]: ";
    for (int_t k = 0, new_graph_top = 0; k < n; k++) {
        if (map[compress.find(k)] == -1)
            map[compress.find(k)] = new_graph_top++;
        map[k] = map[compress.find(k)];
        // std::cerr << map[k] << ' ';
    }
    // std::cerr << std::endl;
    std::vector<adj_list_t> graph(compress.components);
    for (int_t k = 0; k < m; k++) {
        int_t a = map[original_graph[k].a], b = map[original_graph[k].b];
        if (a != b) {
            graph[a].push_back(neighb_t{b, k});
            graph[b].push_back(neighb_t{a, k});
        }
    }
    n = compress.components;
 
    int_t pairs;
    std::cin >> pairs;
    vec start(m, -1);   // for each edge, if it's a -> b, start[] = a.
    // std::cerr << "[debug] requirements:" << std::endl;
    for (int_t k = 0; k < pairs; k++) {
        int_t src, dst;
        std::cin >> src >> dst;
        src = map[src - 1], dst = map[dst - 1];
        // std::cerr << src << ' ' << dst << std::endl;
        if (src == dst)
            continue;
        // TODO: improve BFS
        std::queue<int_t> bfs_queue;
        std::vector<int_t> edge_num(n, -1), prev(n, -1);
        edge_num[src] = INF;    // doesn't exist. Anything not -1 would be good
        bfs_queue.emplace(src);
        while (!bfs_queue.empty()) {
            int_t t = bfs_queue.front();
            bfs_queue.pop();
            if (t == dst)
                break;
            for (const neighb_t& e : graph[t]) {
                if (edge_num[e.neighb] != -1)
                    continue;
                edge_num[e.neighb] = e.edge_idx, prev[e.neighb] = t;
                bfs_queue.push(e.neighb);
            }
        }
        for (int_t pt = dst; pt != src; pt = prev[pt]) {
            assert(edge_num[pt] != -1);
            start[edge_num[pt]] = prev[pt];
        }
    }
 
    std::string ans(m, '@');
    for (int_t k = 0; k < m; k++) {
        if (is_bridge[k]) {
            if (start[k] != -1) {
                if (map[original_graph[k].a] == start[k])
                    ans[k] = 'R';
                else
                    ans[k] = 'L';
            } else {
                ans[k] = 'B';
            }
        } else {
            ans[k] = 'B';
        }
    }
    std::cout << ans << '\n';
}

• Subtask 1: 30 / 30
• Subtask 2: 0 / 30
• Subtask 3: 0 / 40