oj.uz

답안 #815628

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
815628 2023-08-08T17:36:07 Z t6twotwo 사탕 분배 (IOI21_candies) C++17
38 / 100
 5000 ms 27884 KB 
#include "candies.h"
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr ll inf = 1e18;
vector<ll> mn, mn2, mx, mx2, lazy;
void apply_add(int p, ll v) {
    lazy[p] += v;
    mn[p] += v;
    mx[p] += v;
    if (mn2[p] != inf) {
        mn2[p] += v;
    }
    if (mx2[p] != -inf) {
        mx2[p] += v;
    }
}
void apply_upd(int p, ll cmn, ll cmx) {
    if (cmn == cmx) {
        mn[p] = mx[p] = cmn;
        mn2[p] = inf;
        mx2[p] = -inf;
    } else {
        if (cmn < mx[p]) {
            if (mn[p] == mx[p]) {
                mn[p] = cmn;
            }
            if (mn2[p] == mx[p]) {
                mn2[p] = cmn;
            }
            mx[p] = cmn;
        }
        if (cmx > mn[p]) {
            if (mx[p] == mn[p]) {
                mx[p] = cmx;
            }
            if (mx2[p] == mn[p]) {
                mx2[p] = cmx;
            }
            mn[p] = cmx;
        }
    }
}
void push_add(int p) {
    apply_add(p * 2, lazy[p]);
    apply_add(p * 2 + 1, lazy[p]);
    lazy[p] = 0;
}
void push_upd(int p) {
    apply_upd(p * 2, mx[p], mn[p]);
    apply_upd(p * 2 + 1, mx[p], mn[p]);
}
void pull(int p) {
    int l = p * 2;
    int r = p * 2 + 1;
    mx[p] = max(mx[l], mx[r]);
    mn[p] = min(mn[l], mn[r]);
    mx2[p] = max(mx[p] == mx[l] ? mx2[l] : mx[l], mx[p] == mx[r] ? mx2[r] : mx[r]);
    mn2[p] = min(mn[p] == mn[l] ? mn2[l] : mn[l], mn[p] == mn[r] ? mn2[r] : mn[r]);
}
void add(int p, int l, int r, int L, int R, int v) {
    if (R <= l || r <= L) {
        return;
    }
    if (L <= l && r <= R) {
        apply_add(p, v);
        return;
    }
    int m = (l + r + 1) / 2;
    push_add(p);
    push_upd(p);
    add(p * 2, l, m, L, R, v);
    add(p * 2 + 1, m, r, L, R, v);
    pull(p);
}
void upd(int p, int l, int r, int L, int R, ll v, bool f) {
    if (R <= l || r <= L || (!f && v >= mx[p]) || (f && v <= mn[p])) {
        return;
    }
    if (L <= l && r <= R && (mn[p] == mx[p] || (!f && v > mx2[p]) || (f && v < mn2[p]))) {
        if (f) {
            apply_upd(p, inf, v);
        } else {
            apply_upd(p, v, -inf);
        }
        return;
    }
    int m = (l + r + 1) / 2;
    push_add(p);
    push_upd(p);
    upd(p * 2, l, m, L, R, v, f);
    upd(p * 2 + 1, m, r, L, R, v, f);
    pull(p);
}
vector<int> distribute_candies(vector<int> C, vector<int> L, vector<int> R, vector<int> V) {
    for (int &x : R) {
        x++;
    }
    int N = C.size(), Q = V.size();
    if (N <= 2000 && Q <= 2000) {
        vector<int> A(N);
        for (int i = 0; i < Q; i++) {
            for (int j = L[i]; j < R[i]; j++) {
                A[j] += V[i];
                A[j] = max(A[j], 0);
                A[j] = min(A[j], C[j]);
            }
        }
        return A;
    }
    if (*min_element(V.begin(), V.end()) > 0) {
        vector<ll> s(N + 1);
        for (int i = 0; i < Q; i++) {
            s[L[i]] += V[i];
            s[R[i]] -= V[i];
        }
        vector<int> A(N);
        for (int i = 0; i < N; i++) {
            A[i] = min((ll)C[i], s[i]);
            s[i + 1] += s[i];
        }
        return A;
    }
    int M = 2 << __lg(N - 1);
    if (L == vector(Q, 0) && R == vector(Q, N)) {
        vector<int> ord(N);
        iota(ord.begin(), ord.end(), 0);
        sort(ord.begin(), ord.end(), [&](int i, int j) {
            return C[i] < C[j];
        });
        vector<int> st(2 * M), df(2 * M), lz(2 * M, -1);
        vector<ll> sum(2 * M);
        auto pull = [&](int i) {
            st[i] = max(st[i * 2], st[i * 2 + 1]);
            df[i] = max(df[i * 2], df[i * 2 + 1]);
        };
        auto apply = [&](int p, int l, int r, int v, ll s) {
            if (v == 0) {
                st[p] = 0;
                df[p] = C[ord[min(r, N) - 1]];
            } else if (v == 1) {
                st[p] = C[ord[min(r, N) - 1]];
                df[p] = 0;
            }
            st[p] += s;
            df[p] -= s;
            if (v != -1) {
                lz[p] = v;
                sum[p] = s;
            } else {
                sum[p] += s;
            }
        };
        auto push = [&](int p, int l, int r) {
            int m = (l + r + 1) / 2;
            apply(p * 2, l, m, lz[p], sum[p]);
            apply(p * 2 + 1, m, r, lz[p], sum[p]);
            lz[p] = -1;
            sum[p] = 0;
        };
        auto upd = [&](auto upd, int p, int l, int r, int L, int R, int f, int v) -> void {
            if (R <= l || r <= L) {
                return;
            }
            if (L <= l && r <= R) {
                if (f == 0) {
                    apply(p, l, r, 0, 0);
                } else if (f == 1) {
                    apply(p, l, r, 1, 0);
                } else {
                    apply(p, l, r, -1, v);
                }
                return;
            }
            int m = (l + r + 1) / 2;
            push(p, l, r);
            upd(upd, p * 2, l, m, L, R, f, v);
            upd(upd, p * 2 + 1, m, r, L, R, f, v);
            pull(p);
        };
        auto find = [&](auto find, int p, int l, int r, int v, vector<int> &s) -> int {
            if (l + 1 == r) {
                return l;
            }
            int m = (l + r + 1) / 2;
            if (s[p * 2] >= v) {
                return find(find, p * 2, l, m, v, s);
            } else {
                if (s[p * 2 + 1] < v) {
                    while (1) {
                    }
                }
                return find(find, p * 2 + 1, m, r, v, s);
            }
        };
        for (int i = 0; i < N; i++) {
            df[i + M] = C[ord[i]];
            if (i + M >= 2 * M) {
                // while (1) {
                // }
            }
        }
        for (int i = M - 1; i; i--) {
            pull(i);
        }
        for (int i = 0; i < Q; i++) {
            if (V[i] < 0) {
                int f = st[1] < -V[i] ? N : find(find, 1, 0, M, -V[i], st);
                upd(upd, 1, 0, M, 0, f, 0, 0);
                upd(upd, 1, 0, M, f, N, 2, V[i]);
            } else {
                int f = df[1] < V[i] ? N : find(find, 1, 0, M, V[i], df);
                upd(upd, 1, 0, M, 0, f, 1, 0);
                upd(upd, 1, 0, M, f, N, 2, V[i]);
            }
        }
        vector<int> ans(N);
        auto qry = [&](auto qry, int p, int l, int r) -> void {
            if (l + 1 == r) {
                ans[ord[l]] = st[p];
                return;
            }
            int m = (l + r + 1) / 2;
            push(p, l, r);
            qry(qry, p * 2, l, m);
            qry(qry, p * 2 + 1, m, r);
        };
        qry(qry, 1, 0, M);
        return ans;
    }
    mn.resize(2 * M, 0);
    mx.resize(2 * M, 0);
    mn2.resize(2 * M, inf);
    mx2.resize(2 * M, -inf);
    lazy.resize(2 * M);
    for (int i = 0; i < Q; i++) {
        add(1, 0, M, L[i], R[i], V[i]);
        if (V[i] > 0) {
            upd(1, 0, M, L[i], R[i], C[0], 0);
        } else {
            upd(1, 0, M, L[i], R[i], 0, 1);
        }
    }
    for (int i = 1; i < M; i++) {
        push_add(i);
        push_upd(i);
    }
    vector<int> ans(N);
    for (int i = 0; i < N; i++) {
        ans[i] = mx[i + M];
    }
    return ans;
}

Subtask #1
3.0 / 3.0

# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 1 ms 212 KB Output is correct
3 Correct 1 ms 340 KB Output is correct
4 Correct 1 ms 340 KB Output is correct
5 Correct 3 ms 340 KB Output is correct

Subtask #2
8.0 / 8.0

# 결과 실행 시간 메모리 Grader output
1 Correct 83 ms 8808 KB Output is correct
2 Correct 80 ms 8804 KB Output is correct
3 Correct 88 ms 8880 KB Output is correct

Subtask #3
27.0 / 27.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 212 KB Output is correct
2 Correct 170 ms 5728 KB Output is correct
3 Correct 61 ms 24364 KB Output is correct
4 Correct 473 ms 27876 KB Output is correct
5 Correct 538 ms 27880 KB Output is correct
6 Correct 683 ms 27880 KB Output is correct
7 Correct 702 ms 27884 KB Output is correct

Subtask #4
0 / 29.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 212 KB Output is correct
2 Correct 1 ms 212 KB Output is correct
3 Execution timed out 5070 ms 6504 KB Time limit exceeded
4 Halted 0 ms 0 KB -

Subtask #5
0 / 33.0

# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 1 ms 212 KB Output is correct
3 Correct 1 ms 340 KB Output is correct
4 Correct 1 ms 340 KB Output is correct
5 Correct 3 ms 340 KB Output is correct
6 Correct 83 ms 8808 KB Output is correct
7 Correct 80 ms 8804 KB Output is correct
8 Correct 88 ms 8880 KB Output is correct
9 Correct 1 ms 212 KB Output is correct
10 Correct 170 ms 5728 KB Output is correct
11 Correct 61 ms 24364 KB Output is correct
12 Correct 473 ms 27876 KB Output is correct
13 Correct 538 ms 27880 KB Output is correct
14 Correct 683 ms 27880 KB Output is correct
15 Correct 702 ms 27884 KB Output is correct
16 Correct 1 ms 212 KB Output is correct
17 Correct 1 ms 212 KB Output is correct
18 Execution timed out 5070 ms 6504 KB Time limit exceeded
19 Halted 0 ms 0 KB -