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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#91061#6127. Kawa ExamValera_GrinenkoWA 2ms20612kbC++147.6kb2023-03-26 22:31:182023-03-26 22:31:19

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-03-26 22:31:19]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:20612kb
  • [2023-03-26 22:31:18]
  • 提交

answer

//#pragma GCC optimize("Ofast", "unroll-loops")
//#pragma GCC target("sse", "sse2", "sse3", "ssse3", "sse4")

#ifdef __APPLE__

#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <cstdint>
#include <cstring>
#include <string>
#include <cstdlib>
#include <vector>
#include <bitset>
#include <map>
#include <queue>
#include <ctime>
#include <stack>
#include <set>
#include <list>
#include <random>
#include <deque>
#include <functional>
#include <iomanip>
#include <sstream>
#include <fstream>
#include <complex>
#include <numeric>
#include <immintrin.h>
#include <cassert>
#include <array>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <thread>

#else
#include <bits/stdc++.h>
#endif

#define all(a) a.begin(),a.end()
#define len(a) (int)(a.size())
#define mp make_pair
#define pb push_back
#define fir first
#define sec second
#define fi first
#define se second

using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef long double ld;

template<typename T>
bool umin(T &a, T b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}

template<typename T>
bool umax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

#if __APPLE__
#define D for (bool _FLAG = true; _FLAG; _FLAG = false)
#define LOG(...) print(#__VA_ARGS__" ::", __VA_ARGS__) << endl

template<class ...Ts>
auto &print(Ts ...ts) { return ((cerr << ts << " "), ...); }

#else
#define D while (false)
#define LOG(...)
#endif

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int max_n = 2e5 + 42;

vector<pair<int, int> > g[max_n], g2[max_n];

vector<int> ans, two_edge_component, comp_num;
vector<vector<int> > two_edge_components;
int cur_component = 0;

vector<bool> used, is_bridge;

int a[max_n];

map<int, int> am_dfs1;

vector<int> nodes_dfs1;

void dfs1(int v, bool save_nodes = false) {
    if(save_nodes) nodes_dfs1.pb(v);
    used[v] = true;
    am_dfs1[a[v]]++;
    for(auto& to : g[v])
        if(!used[to.fi])
            dfs1(to.fi, save_nodes);
}

vector<bool> visited;
vector<int> tin, low;
int timer;

void dfs2(int v, int p = -1) {
    visited[v] = true;
    tin[v] = low[v] = timer++;
    for (auto& to : g[v]) {
        if (to.se == p) continue;
        if (visited[to.fi]) {
            low[v] = min(low[v], tin[to.fi]);
        } else {
            dfs2(to.fi, to.se);
            low[v] = min(low[v], low[to.fi]);
            if (low[to.fi] > tin[v]) is_bridge[to.se] = true;
        }
    }
}

void find_bridges(int n) {
    timer = 0;
    visited.assign(n, false);
    tin.assign(n, -1);
    low.assign(n, -1);
    for (int i = 0; i < n; ++i) {
        if (!visited[i])
            dfs2(i);
    }
}

void dfs3(int v) {
    comp_num[v] = cur_component;
    two_edge_component.pb(v);
    used[v] = true;
    for(auto& to : g[v])
        if(!used[to.fi] && !is_bridge[to.se])
            dfs3(to.fi);
}

struct kek {
    //add integer x to multiset
    //delete integer x from multiset
    //get number of occurences of mode(most frequent element) in multiset
    int am_element[max_n] = {}; //number of elements equal to x
    int am_am_element[max_n] = {}; //number of elements that have x occurences in multiset
    int res = 0;
    void init() {
        res = 0;
        for(int i = 0; i < max_n; i++) {
            am_element[i] = am_am_element[i] = 0;
        }
        am_am_element[0] = max_n + 42;
    }
    void add(int val) {
        am_am_element[am_element[val]]--;
        am_element[val]++;
        am_am_element[am_element[val]]++;
        umax(res, am_element[val]);
    }
    void del(int val) {
        am_am_element[am_element[val]]--;
        am_element[val]--;
        am_am_element[am_element[val]]++;
        if(res && !am_am_element[res]) res--;
    }
    int get() {
        return res;
    }
};

kek kek1, kek2; //kek1 -> inside of subtree, kek2 -> outside of subtree

vector<int> sz;

void dfssz(int v, int p) {
    sz[v] = len(two_edge_components[v]);
    for(auto& to : g2[v])
        if(to.fi != p) {
            dfssz(to.fi, v);
            sz[v] += sz[to.fi];
        }
}

vector<int> vec[max_n];

int base_ans = 0;

void dfs4(int v, int p, bool keep) {
    int par_edge = -1;
    int mx_child = -1, child = -1;
    for(auto& to : g2[v]) {
        if(to.fi == p) { par_edge = to.se; continue; }
        if(umax(mx_child, sz[to.fi])) {
            child = to.fi;
        }
    }
    for(auto& to : g2[v])
        if(to.fi != p && to.fi != child)
            dfs4(to.fi, v, 0);
    if(child != -1) {
        dfs4(child, v, 1);
        vec[v].swap(vec[child]);
    } else vec[v] = vector<int>();
    for(auto& node : two_edge_components[v]) {
        vec[v].pb(node);
        kek1.add(a[node]);
        kek2.del(a[node]);
    }
    for(auto& to : g2[v])
        if(to.fi != p && to.fi != child)
            for(auto& node : vec[to.fi]) {
                vec[v].pb(node);
                kek1.add(a[node]);
                kek2.del(a[node]);
            }
    if(par_edge != -1) {
        ans[par_edge] = kek1.get() + kek2.get() + base_ans;
    }
    if(!keep) {
        for(auto& node : vec[v]) {
            kek1.del(a[node]);
            kek2.add(a[node]);
        }
    }
}

void solve() {
    int n, m;
    cin >> n >> m;
    for(int i = 0; i < n; i++) {
        g[i].clear();
        g2[i].clear();
        cin >> a[i];
    }
    vector<pair<int, int> > edges;
    for(int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        u--; v--;
        edges.pb({u, v});
        g[u].pb({v, i});
        g[v].pb({u, i});
    }
    vector<int> roots;
    used.assign(n, false);
    base_ans = 0;
    for(int i = 0; i < n; i++)
        if(!used[i]) {
            roots.pb(i);
            am_dfs1.clear();
            dfs1(i);
            int mx = 0;
            for(auto& x : am_dfs1) umax(mx, x.se);
            base_ans += mx;
        }
    ans.assign(m, base_ans);
    is_bridge.assign(m, false);
    find_bridges(n);
    used.assign(n, false);
    two_edge_component.assign(n, -1);
    cur_component = 0;
    two_edge_components.clear();
    comp_num.assign(n, -1);
    for(int i = 0; i < n; i++)
        if(!used[i]) {
            two_edge_component.clear();
            dfs3(i);
            two_edge_components.pb(two_edge_component);
            cur_component++;
        }
    for(int i = 0; i < m; i++)
        if(comp_num[edges[i].fi] != comp_num[edges[i].se]) {
            g2[comp_num[edges[i].fi]].pb({comp_num[edges[i].se], i});
            g2[comp_num[edges[i].se]].pb({comp_num[edges[i].fi], i});
        }
    used.assign(n, false);
    sz.assign(n, 0);
    for(auto& root : roots) {
        am_dfs1.clear();
        dfs1(root, true);
        int mx = 0;
        for(auto& x : am_dfs1) umax(mx, x.se);
        for(auto& v : nodes_dfs1) kek2.add(a[v]);
        base_ans -= mx;
        dfssz(comp_num[root], comp_num[root]);
        dfs4(comp_num[root], comp_num[root], false);
        base_ans += mx;
        for(auto& v : nodes_dfs1) kek2.del(a[v]);
        nodes_dfs1.clear();
    }
    for(auto& x : ans) cout << x << ' ';
    cout << '\n';
}

signed main() {
//   freopen("input.txt", "r", stdin);
//   freopen("output.txt", "w", stdout);

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    kek1.init(); kek2.init();

    int t = 1;

    cin >> t;

    while (t--) solve();

}

/*
KIVI
 testing
*/

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 0
Wrong Answer
time: 2ms
memory: 20612kb

input:

3
7 5
1 2 1 2 1 2 1
1 2
1 3
2 4
5 6
5 7
3 3
1 2 3
1 2
1 3
2 3
2 3
12345 54321
1 2
1 2
1 1

output:

6 5 5 5 4 
1 1 1 
1 1 1 

result:

wrong answer 1st lines differ - expected: '6 5 5 5 4', found: '6 5 5 5 4 '