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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#853523#9731. Fuzzy Rankingucup-team5234#WA 0ms3628kbC++235.2kb2025-01-11 17:20:492025-01-11 17:20:51

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你现在查看的是最新测评结果

  • [2025-01-11 17:20:51]
  • 评测
  • 测评结果:WA
  • 用时:0ms
  • 内存:3628kb
  • [2025-01-11 17:20:49]
  • 提交

answer

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

namespace atcoder {
namespace internal {

template <class E> struct csr {
    std::vector<int> start;
    std::vector<E> elist;
    explicit csr(int n, const std::vector<std::pair<int, E>>& edges)
        : start(n + 1), elist(edges.size()) {
        for (auto e : edges) {
            start[e.first + 1]++;
        }
        for (int i = 1; i <= n; i++) {
            start[i] += start[i - 1];
        }
        auto counter = start;
        for (auto e : edges) {
            elist[counter[e.first]++] = e.second;
        }
    }
};

}  // namespace internal

}  // namespace atcoder


namespace atcoder {
namespace internal {

// Reference:
// R. Tarjan,
// Depth-First Search and Linear Graph Algorithms
struct scc_graph {
  public:
    explicit scc_graph(int n) : _n(n) {}

    int num_vertices() { return _n; }

    void add_edge(int from, int to) { edges.push_back({from, {to}}); }

    // @return pair of (# of scc, scc id)
    std::pair<int, std::vector<int>> scc_ids() {
        auto g = csr<edge>(_n, edges);
        int now_ord = 0, group_num = 0;
        std::vector<int> visited, low(_n), ord(_n, -1), ids(_n);
        visited.reserve(_n);
        auto dfs = [&](auto self, int v) -> void {
            low[v] = ord[v] = now_ord++;
            visited.push_back(v);
            for (int i = g.start[v]; i < g.start[v + 1]; i++) {
                auto to = g.elist[i].to;
                if (ord[to] == -1) {
                    self(self, to);
                    low[v] = std::min(low[v], low[to]);
                } else {
                    low[v] = std::min(low[v], ord[to]);
                }
            }
            if (low[v] == ord[v]) {
                while (true) {
                    int u = visited.back();
                    visited.pop_back();
                    ord[u] = _n;
                    ids[u] = group_num;
                    if (u == v) break;
                }
                group_num++;
            }
        };
        for (int i = 0; i < _n; i++) {
            if (ord[i] == -1) dfs(dfs, i);
        }
        for (auto& x : ids) {
            x = group_num - 1 - x;
        }
        return {group_num, ids};
    }

    std::vector<std::vector<int>> scc() {
        auto ids = scc_ids();
        int group_num = ids.first;
        std::vector<int> counts(group_num);
        for (auto x : ids.second) counts[x]++;
        std::vector<std::vector<int>> groups(ids.first);
        for (int i = 0; i < group_num; i++) {
            groups[i].reserve(counts[i]);
        }
        for (int i = 0; i < _n; i++) {
            groups[ids.second[i]].push_back(i);
        }
        return groups;
    }

  private:
    int _n;
    struct edge {
        int to;
    };
    std::vector<std::pair<int, edge>> edges;
};

}  // namespace internal

}  // namespace atcoder

namespace atcoder {

struct scc_graph {
  public:
    scc_graph() : internal(0) {}
    explicit scc_graph(int n) : internal(n) {}

    void add_edge(int from, int to) {
        int n = internal.num_vertices();
        assert(0 <= from && from < n);
        assert(0 <= to && to < n);
        internal.add_edge(from, to);
    }

    std::vector<std::vector<int>> scc() { return internal.scc(); }

  private:
    internal::scc_graph internal;
};

}  // namespace atcoder

using namespace atcoder;

void solve(){
    int N,K,Q;
    cin >> N >> K >> Q;
    scc_graph scc(N);
    vector<vector<int>> V(K, vector<int>(N));
    for(int i = 0; i < K; i++){
        for(int j = 0; j < N; j++){
            cin >> V[i][j];
            V[i][j]--;
            if(j > 0){
                scc.add_edge(V[i][j-1], V[i][j]);
            }
        }
    }
    auto ret = scc.scc();
    vector<int> rev(N);
    for(int i = 0; i < ret.size(); i++){
        for(auto v: ret[i]){
            rev[v] = i;
        }
    }
    vector<ll> c(ret.size() + 1);
    vector<ll> sum(ret.size() + 1);
    for(int i = 0; i < ret.size(); i++){
        c[i + 1] = c[i] + ret[i].size();
        sum[i + 1] = sum[i] + ll(ret[i].size()) * ll(ll(ret[i].size()) - 1) / 2;
    }
    auto f = [&](int l, int r) -> ll {
        if(r < l)return 0;
        return sum[r] - sum[l];
    };
    auto cnt = [&](int l, int r = -2) -> ll {
        if(r == -2) r = l + 1;
        if(r < l)return 0;
        return c[r] - c[l];
    };

    ll v=0;
    while(Q--){
        ll id, l, r;
        cin >> id >> l >> r;
        id += v;
        id %= K;
        l += v;
        l %= N;
        r += v;
        r %= N;
        if(rev[l] == rev[r]){
            v = (r - l + 1) * (r - l) / 2;
            cout << v << "\n";
        }else{
            ll C = cnt(0, rev[l]);
            ll Cnt = cnt(rev[l]) - (l - C);
            v = 0;
            v += Cnt * (Cnt - 1) / 2;
            C = cnt(0, rev[r]);
            Cnt = r - C + 1;
            v += Cnt * (Cnt - 1) / 2;
            v += f(rev[l], rev[r] + 1 - 1);
            cout << v << "\n";
        }
    }
}


signed main(){
    int T;
    cin >> T;
    while(T--){
        solve();
    }
}

详细

Test #1:

score: 0
Wrong Answer
time: 0ms
memory: 3628kb

input:

2
5 2 2
1 2 3 4 5
5 4 3 2 1
1 0 2
1 2 1
5 3 3
1 2 3 4 5
1 3 2 4 5
1 2 3 5 4
0 0 2
0 2 3
1 0 3

output:

3
10
1
1
3

result:

wrong answer 5th lines differ - expected: '2', found: '3'