제출 #516106

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
516106		goodluck2020	Financial Report (JOI21_financial)	C++14	100 / 100	427 ms	29676 KiB

Main

#include <bits/stdc++.h>
#define task "JOI21_FINANCIAL"

using namespace std;
const int N = 3e5 + 5;
int n, A[N], D;
set < int > Set;
int CP[N], cnt;
void compress()
{
    for(int i = 1; i <= n; i++) Set.insert(A[i]);
    for(int j : Set) CP[++cnt] = j;
    for(int i = 1; i <= n; i++)
        A[i] = lower_bound(CP + 1, CP + cnt + 1, A[i]) - CP;
}
namespace sub2
{
    int dp[405][405], ans;
    void solve()
    {
        for(int i = 0; i <= n; i++)
        for(int j = 0; j <= n; j++) dp[i][j] = -1e9;
        dp[0][0] = 0;
        for(int i = 0; i < n; i++)
        {
            if(i)
            dp[i][A[i]] = max(dp[i][A[i]], 1);
            for(int j = 0; j <= n; j++)
                for(int t = i + 1; t <= i + D; t++)
                    if(t <= n)
                    dp[t][max(A[t], j)] = max(dp[t][max(A[t], j)], dp[i][j] + (A[t] > j));
        }
        for(int j = 0; j <= n; j++) ans = max(ans, dp[n][j]);
        cout << ans;
    }
}
namespace sub3
{
    int dp[7005], ans;
    void solve()
    {
        for(int i = 1; i <= n; i++)
        {
            dp[i] = 1;
            bool ok = true;
            int cnt = 0;
            for(int j = i - 1; j >= 1; j--)
            {
                if(ok && A[j] < A[i]) dp[i] = max(dp[i], dp[j] + 1);
                if(A[j] >= A[i]) cnt++;
                else cnt = 0;
                if(cnt >= D) ok = false;
            }
            ans = max(ans, dp[i]);
        }
        cout << ans;
    }
}
namespace sub4
{
    int L[N], St[N], top, ans;
    priority_queue < int > pq;
    void solve()
    {
        for(int i = 1; i <= n; i++)
        {
            while(top && A[i] > A[St[top]]) top--;
            L[i] = St[top] + 1;
            St[++top] = i;
        }
        for(int i = n; i >= 1; i--)
        {
            pq.push(L[i]);
            while(pq.top() > i) pq.pop();
            ans = max(ans, (int)pq.size());
        }
        cout << ans;
    }
}
namespace sub5
{
    int ans;
    struct FenwickTree
    {
        int BIT[N];
        int get(int p)
        {
            int res = 0;
            for( ; p ; p -= p & -p)
                res = max(res, BIT[p]);
            return res;
        }
        void update(int p, int val)
        {
            for( ; p <= 3e5; p += p & -p)
                BIT[p] = max(BIT[p], val);
        }
    } FT;
    void solve()
    {
        for(int i = 1; i <= n; i++)
        {
            int tmp = FT.get(A[i] - 1) + 1;
            FT.update(A[i], tmp);
            ans = max(ans, tmp);
        }
        cout << ans;
    }
}
namespace sub6
{
    int Min[N], R[N];
    void BuildMin()
    {
        stack < int > st;
        st.push(n+1);
        for(int i = n; i >= 1; i--)
        {
            while(st.size()>1 && A[st.top()]>A[i]) st.pop();
            R[i] = st.top() - 1;
            st.push(i);
        }
        int p = 1;
        for(int i = 1; i <= n - D + 1; i++)
        {
            int j = i + D - 1;
            while(!(i <= p && p <= j && R[p] >= j) && p<=n) p++;
            Min[i] = A[p];
        }
        ///Min[i] la GTNN trong doan [i; i + D - 1]
    }
    struct FenwickTree
    {
        int BIT[N];
        int get(int p)
        {
            int res = 0;
            for( ; p <= 3e5; p += p & -p)
                res = max(res, BIT[p]);
            return res;
        }
        void update(int p, int val)
        {
             for( ; p ; p -= p & -p)
                BIT[p] = max(BIT[p], val);
        }
    } FT;
    struct SegmentTree
    {
        int ST[4*N];
        int get(int id, int l, int r, int u, int v)
        {
            if(l > v || r < u) return 0;
            if(u <= l && r <= v) return ST[id];
            int mid = (l + r)/2;
            return max(get(id*2, l, mid, u, v), get(id*2+1, mid + 1, r, u, v));
        }
        void update(int id, int l, int r, int x, int val)
        {
            if(l > x || r < x) return;
            if(l == r)
            {
                ST[id] = max(ST[id], val);
                return;
            }
            int mid = (l + r)/2;
            update(id*2, l, mid, x, val);
            update(id*2 + 1, mid + 1, r, x, val);
            ST[id] = max(ST[id*2], ST[id*2 + 1]);
        }
    } IT;
    int leftmost[N], ans = 0, ord[N];
    void solve()
    {
        BuildMin();
        for(int i = 1; i <= n; i++)
        {
            leftmost[i] = FT.get(A[i]) + 1;
            ///True in [pos, i - 1]
            if(i >= D) FT.update(Min[i - D + 1], i - D + 1);
        }
        for(int i = 1; i <= n; i++) ord[i] = i;
        sort(ord + 1, ord + n + 1, [&] (int x, int y){
             if(A[x] == A[y]) return x > y; else return A[x] < A[y];
             });
        for(int j = 1; j <= n; j++)
        {
            int i = ord[j];
            int tmp = IT.get(1, 1, n, leftmost[i], i - 1) + 1;
            ans = max(ans, tmp);
            IT.update(1, 1, n, i, tmp);
        }
        cout << ans;
    }
}
int main()
{
    if(fopen(task ".inp","r"))
    {
        freopen(task ".inp","r",stdin);
        freopen(task ".out","w",stdout);
    }
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cin >> n >> D;
    for(int i = 1; i <= n; i++) cin >> A[i];
    compress();
    if(n <= 400) sub2::solve();
    else if(n <= 7000) sub3::solve();
    else if(D == 1) sub4::solve();
    else if(D == n) sub5::solve();
    else sub6::solve();
}

컴파일 시 표준 에러 (stderr) 메시지

Main.cpp: In function 'int main()':
Main.cpp:200:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  200 |         freopen(task ".inp","r",stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~
Main.cpp:201:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  201 |         freopen(task ".out","w",stdout);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 20 / 20
• Subtask 4: 12 / 12
• Subtask 5: 5 / 5
• Subtask 6: 35 / 35