제출 #1251846

#제출 시각아이디문제언어결과실행 시간메모리
1251846s4dz3개의 봉우리 (IOI25_triples)C++20
85 / 100
2093 ms24688 KiB
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
//sub4:
vector<int> construct_range(int M, int K) {
    vector<int> ans1 = {
        4,  3,  1,  2,  1,
        4,  3,  2,  7,  6,
        5,  8, 11, 10,  9,
        1,  7,  2,  3,  4,
    };

    if(M <= 20){
        while(ans1.size() > M || ans1.back() >= M) ans1.pop_back();
        return ans1;
    }
    vector<int> S = {0};
    vector<bool> in_span(2*M, false);
    int span_cnt = 0;
    vector<pair<int,int>> pts;

    while(span_cnt < M-1) {
        int best_t = -1, best_gain = -1;
        vector<bool> inS(2*M, false);
        for(int u:S) inS[u] = true;
        for(int t = 0; t <= 2*(M-1); t += 2) if(!inS[t]) {
            int gain = 0;
            for(int u:S) {
                if(u==t) continue;
                int s = u + t;
                if(s >= 2 && s <= 2*(M-1) && (s%2==0) && !in_span[s]) ++gain;
            }
            if(gain > best_gain) {
                best_gain = gain;
                best_t = t;
            }
        }
        if(best_t < 0) break;
        for(int u:S) {
            if(u == best_t) continue;
            int s = u + best_t;
            if(s >= 2 && s <= 2*(M-1) && (s%2==0) && !in_span[s]) {
                int x = min(u, best_t), y = max(u, best_t);
                pts.emplace_back(x, y);
                in_span[s] = true;
                ++span_cnt;
            }
        }
        S.push_back(best_t);
    }
    vector<int> H(M, 1);
    for(auto [x,y] : pts) {
        int idx = (x + y) / 2;
        int h   = (y - x) / 2;
        if(idx >= 0 && idx < M) {
            H[idx] = h;
        }
    }
    return H;
}
/*long long count_triples(vector<int> H)
{
    int n = H.size();
    ll ans = 0;
    for (int k = 0; k < n; k++)
    {
        int c = H[k];
        int i = k - c;
        if (i < 0 || i >= n) continue;

        int a = H[i];
        if (a > c - a) continue;

        int need = c - a;
        int j1   = i + a;
        int j2   = k - a;
        if (i < j1 && j1 < k && H[j1] == need) ans++;
        if (j2 != j1 && i < j2 && j2 < k && H[j2] == need) ans++;
    }
    return ans;
}*/

const int nmax = 2e5 + 5;
unordered_map <int, vector <int>> m;
int n;
ll count_triples(vector <int> H)
{
    n = H.size();
    ll cnt = 0;
    for (int j=0; j<n; j++)
    {
        int l = j - H[j];
        if(l >= 0)
        {
            int x = H[l];
            int y = H[j] - x;
            if(y > 0)
            {
                if(H[l + x] == y) cnt++;
                if(y != x && H[l + y] == y) cnt++;
            }
        }
        int r = j + H[j];
        if(r < n)
        {
            int x = H[r];
            int y = H[j] - x;
            if(y > 0)
            {
                if(H[j + x] == y) cnt++;
                if(y != x && H[j + y] == y) cnt++;
            }
        }
    }
    for (int k = 0; k < n; ++k)
    {
        int hk = H[k];
        for (int i : m[k - hk])
        {
            int j1 = i+H[i];
            int j2 = i+H[k];
            if (j1<k && H[j1]==k-i) cnt++;
            if (j2<k && j2!=j1 && H[j2]==k-i) cnt++;
        }

        m[k + H[k]].push_back(k);
    }
    return cnt;
}
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