Submission #571102

#TimeUsernameProblemLanguageResultExecution timeMemory
571102elazarkorenMeetings (IOI18_meetings)C++17
0 / 100
20 ms1616 KiB
#include "meetings.h"
#include <bits/stdc++.h>
#define x first
#define y second
#define all(v) v.begin(), v.end()
#define chkmin(a, b) a = min(a, b)
#define chkmax(a, b) a = max(a, b)
using namespace std;
typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pii;
typedef vector<pii> vii;

const ll infinity = 1e18;

struct Seg {
    int n;
    vi seg;
    Seg (int m) {
        for (n = 1; n < m; n <<= 1);
        seg.resize(2 * n);
    }
    void update(int i, int x) {
        seg[i += n] = x;
        for (i >>= 1; i; i >>= 1) {
            seg[i] = max(seg[i << 1], seg[i << 1 | 1]);
        }
    }
    int query(int l, int r) {
        int ans = 0;
        for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
            if (l & 1) ans += seg[l++];
            if (r & 1) ans += seg[--r];
        }
        return ans;
    }
};

vi minimum_costs(vector<int> h, vector<int> L, vector<int> R) {
    int q = L.size(), n = h.size();
    vi ans(q);
    vector<int> left(n), right(n);
    left[0] = h[0] == 1 ? -1 : 0;
    for (int i = 1; i < n; i++) {
        left[i] = left[i - 1];
        if (h[i] == 2) left[i] = i;
    }
    right[n - 1] = h[n - 1] == 1 ? n : n - 1;
    for (int i = n - 2; i >= 0; i--) {
        right[i] = right[i + 1];
        if (h[i] == 2) right[i] = i;
    }
    Seg seg(n);
    for (int i = 0; i < n; i++) {
        if (h[i] == 1) {
            seg.update(i, right[i] - left[i] - 1);
        }
    }
    for (int j = 0; j < q; j++) {
        int l = L[j], r = R[j];
        if (h[l] == 1 && right[l] > r) {
            ans[j] = r - l + 1;
            continue;
        }
        int max_len = 0;
        if (h[l] == 1) {
            chkmax(max_len, right[l] - l);
            l = right[l];
        }
        if (h[r] == 1) {
            chkmax(max_len, r - left[r]);
            r = left[r];
        }
        chkmax(max_len, seg.query(l, r));
        ans[j] = 2 * (R[j] - L[j] + 1) - max_len;
    }
    return ans;
}
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