답안 #936116

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
936116 2024-03-01T07:09:42 Z GrindMachine 늑대인간 (IOI18_werewolf) C++17
34 / 100
1101 ms 74008 KB
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 2e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "werewolf.h"

template<typename T>
struct sparse_table1 {
    /*============================*/

    T merge(T a, T b) {
        return min(a,b);
    }

    /*============================*/

    vector<vector<T>> table;
    vector<int> bin_log;
    int LOG = 0;

    sparse_table1() {

    }

    sparse_table1(vector<T> &a, int n) {
        while ((1 << LOG) <= n) LOG++;

        table = vector<vector<T>>(n, vector<T>(LOG));
        bin_log = vector<int>(n + 1);

        rep(i, n) table[i][0] = a[i];

        rep1(j, LOG - 1) {
            rep(i, n) {
                int jump = 1 << (j - 1);
                if (i + jump >= n) {
                    break;
                }

                table[i][j] = merge(table[i][j - 1], table[i + jump][j - 1]);
            }
        }

        bin_log[1] = 0;
        for (int i = 2; i <= n; ++i) {
            bin_log[i] = bin_log[i / 2] + 1;
        }
    }

    T query(int l, int r) {
        int len = r - l + 1;
        int k = bin_log[len];

        T val1 = table[l][k];
        T val2 = table[r - (1 << k) + 1][k];

        return merge(val1, val2);
    }
};

template<typename T>
struct sparse_table2 {
    /*============================*/

    T merge(T a, T b) {
        return max(a,b);
    }

    /*============================*/

    vector<vector<T>> table;
    vector<int> bin_log;
    int LOG = 0;

    sparse_table2() {

    }

    sparse_table2(vector<T> &a, int n) {
        while ((1 << LOG) <= n) LOG++;

        table = vector<vector<T>>(n, vector<T>(LOG));
        bin_log = vector<int>(n + 1);

        rep(i, n) table[i][0] = a[i];

        rep1(j, LOG - 1) {
            rep(i, n) {
                int jump = 1 << (j - 1);
                if (i + jump >= n) {
                    break;
                }

                table[i][j] = merge(table[i][j - 1], table[i + jump][j - 1]);
            }
        }

        bin_log[1] = 0;
        for (int i = 2; i <= n; ++i) {
            bin_log[i] = bin_log[i / 2] + 1;
        }
    }

    T query(int l, int r) {
        int len = r - l + 1;
        int k = bin_log[len];

        T val1 = table[l][k];
        T val2 = table[r - (1 << k) + 1][k];

        return merge(val1, val2);
    }
};

vector<int> adj[N];

std::vector<int> check_validity(int n, std::vector<int> X, std::vector<int> Y,
                                std::vector<int> S, std::vector<int> T,
                                std::vector<int> L, std::vector<int> R) {

    int m = sz(X), q = sz(S);
    rep(i,m){
        int u = X[i], v = Y[i];
        adj[u].pb(v), adj[v].pb(u);
    }

    int root = -1;
    rep(i,n){
        if(sz(adj[i]) == 1){
            root = i;
        }
    }

    assert(root != -1);

    vector<int> order;
    int curr = root, par = -1;

    while(true){
        order.pb(curr);
        int nxt = -1;
        trav(v,adj[curr]){
            if(v != par){
                nxt = v;
            }
        }
        if(nxt == -1) break;
        par = curr;
        curr = nxt;
    }

    vector<int> pos(n);
    rep(i,n) pos[order[i]] = i;

    sparse_table1<int> st_min(order,n);
    sparse_table2<int> st_max(order,n);

    vector<int> ans(q);

    rep(i,q){
        int s = S[i], t = T[i], lx = L[i], rx = R[i];
        int ls = -1, rs = -1, lt = -1, rt = -1;

        {
            int lo = 0, hi = pos[s];
            while(lo <= hi){
                int mid = (lo+hi) >> 1;
                if(st_min.query(mid,pos[s]) >= lx){
                    ls = mid;
                    hi = mid-1;
                }
                else{
                    lo = mid+1;
                }
            }
        }

        {
            int lo = pos[s], hi = n-1;
            while(lo <= hi){
                int mid = (lo+hi) >> 1;
                if(st_min.query(pos[s],mid) >= lx){
                    rs = mid;
                    lo = mid+1;
                }
                else{
                    hi = mid-1;
                }
            }
        }


        {
            int lo = 0, hi = pos[t];
            while(lo <= hi){
                int mid = (lo+hi) >> 1;
                if(st_max.query(mid,pos[t]) <= rx){
                    lt = mid;
                    hi = mid-1;
                }
                else{
                    lo = mid+1;
                }
            }
        }

        {
            int lo = pos[t], hi = n-1;
            while(lo <= hi){
                int mid = (lo+hi) >> 1;
                if(st_max.query(pos[t],mid) <= rx){
                    rt = mid;
                    lo = mid+1;
                }
                else{
                    hi = mid-1;
                }
            }
        }

        // do [ls,rs] and [lt,rt] have at least 1 point in common?
        if(min(rs,rt) >= max(ls,lt)){
            ans[i] = 1;
        }
    }

    return ans;
}
# 결과 실행 시간 메모리 Grader output
1 Incorrect 3 ms 4952 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 3 ms 4952 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 820 ms 73848 KB Output is correct
2 Correct 1094 ms 73876 KB Output is correct
3 Correct 995 ms 73872 KB Output is correct
4 Correct 1022 ms 73876 KB Output is correct
5 Correct 1055 ms 73868 KB Output is correct
6 Correct 862 ms 73924 KB Output is correct
7 Correct 972 ms 74008 KB Output is correct
8 Correct 803 ms 73864 KB Output is correct
9 Correct 443 ms 73980 KB Output is correct
10 Correct 346 ms 73860 KB Output is correct
11 Correct 385 ms 73816 KB Output is correct
12 Correct 333 ms 73640 KB Output is correct
13 Correct 1098 ms 73872 KB Output is correct
14 Correct 1006 ms 73928 KB Output is correct
15 Correct 1018 ms 73876 KB Output is correct
16 Correct 1101 ms 73872 KB Output is correct
17 Correct 1014 ms 73876 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Incorrect 3 ms 4952 KB Output isn't correct
2 Halted 0 ms 0 KB -