2019 Multi-University Training Contest 2

1002.Beauty Of Unimodal Sequence

题意：给定一个数组，找出其最长，且字典序最小/大的单峰序列。

我的思路：对于字典序最小，收先需处理$i-n$的包括i数字的最长下降和最长单峰子序列，再从前往后贪心选数且判断该选择是否合法。最长下降可以用以前经典方法，最长单峰可以用线段树维护$l$长单峰序列最后一个值。字典序最大直接反转数组再求一遍即可。

代码：

#include <bits/stdc++.h>
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;

const int N = 3e5 + 5;

int nod[N<<2];
void build(int l, int r, int rt) {
  nod[rt] = 0;
  if (l == r) {
    return ;
  }
  int mid = (l+r)>>1;
  build(lson);
  build(rson);
}
void update(int p, int x, int l, int r, int rt) {
  if (l == r) {
    if (nod[rt] < x) {
      nod[rt] = x;
    }
    return ;
  }
  int mid = (l+r)>>1;
  if (p <= mid) update(p, x, lson);
  else update(p, x, rson);
  nod[rt] = max(nod[rt<<1], nod[rt<<1|1]);
}
int query(int x, int l, int r, int rt) {
  if (l == r) {
    return l;
  }
  int mid = (l+r)>>1;
  if (nod[rt<<1|1] > x) return query(x, rson);
  else return query(x, lson);
}

int a[N];
int c[N];
int suf1[N], suf2[N];
int ans[N];

void solve(int n, int f)
{
  build(1, n, 1);
  int l1, l2;
  int ansl = 0;
  l1 = l2 = c[0] = 0;
  for (int i = n - 1; i >= 0; -- i) {
    int k = lower_bound(c, c + l1 + 1, a[i]) - c;
    c[k] = a[i];
    suf1[i] = k;
    l1 = max(k, l1);
    int ps = 1;
    if (nod[1] > a[i]) {
      ps = query(a[i], 1, n, 1) + 1;
    }
    update(ps, a[i], 1, n, 1);
    update(l1, a[i], 1, n, 1);
    suf2[i] = max(ps, l1);
    ansl = max(ansl, suf2[i]);
    //cout << i << " " << suf1[i] << " " << suf2[i] << endl;
  }
  int F = 0; //f = 0 ins : 1 des
  int p = 0;
  for (int i = 0; i < n; ++ i) {
    if (p == 0) {
      if (suf2[i] == ansl)
        ans[p ++] = i;
      continue;
    }
    if (!F && a[i] > a[ans[p - 1]] && p + suf2[i] == ansl) {
      ans[p ++] = i;
      continue;
    }
    if (a[i] < a[ans[p - 1]] && p + suf1[i] == ansl) {
      ans[p ++] = i; F = 1;
      continue;
    }
  }
  for (int i = 0; i < p; ++ i) {
    int pt = n - ans[p - i - 1];
    if (f) pt = ans[i] + 1;
    printf("%d%c", pt, i == p-1 ? '\n' : ' ');
  }
}

int main()
{
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; ++ i) {
      scanf("%d", a + i);
    }
    solve(n, 1);
    for (int i = 0; i < n / 2; ++ i) {
      swap(a[i], a[n - i - 1]);
    }
    solve(n, 0);
  }
  return 0;
}

1012.Longest Subarray

题意：给定一个数组，找出最长连续子序列使得每种数出现次数均大于$K$。

我的思路：固定右端点，按照数字种类维护每种数字合法左端点，对这些位置$+1$，查询时找出最左边等于$C$的位置即可。

代码：

#include <bits/stdc++.h>
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;

namespace fastIO {
  //fread -> read
  const int BUF_SIZE = 1000000;
  bool IOerror = 0;
  inline char nc() {
    //FILE* fp=fopen("123.txt","r");
    static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
    if(p1 == pend) {
      p1 = buf;
      pend = buf + fread(buf, 1, BUF_SIZE, stdin);
      if(pend == p1) {
        IOerror = 1;
        return -1;
      }
    }
    return *p1++;
  }

  //输入挂,EOF返回0,有输入返回1.
  template <class T>
  inline bool read(T &ret) {
    char c;
    if (c=nc(),c==EOF)return 0; //EOF
    while(c!='-'&&(c<'0'||c>'9'))c=nc();
    int sgn =(c=='-')?-1:1;
    ret=(c=='-')?0:(c - '0');
    while(c=nc(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    ret *= sgn;
    return 1;
  }
  template <class T>
  inline void print(T x) {
    if(x>9) print(x/10);
    putchar(x%10+'0');
  }
}; using fastIO::read;

const int N = 1e5 + 5;

int seg[N<<2];
int c[N<<2];

void build(int l, int r, int rt)
{
  seg[rt] = c[rt] = 0;
  if (l == r) {
    return ;
  }
  int mid = (l+r) >> 1;
  build(lson);
  build(rson);
}

void pushdown(int rt)
{
  if (c[rt]) {
    seg[rt<<1] += c[rt];
    seg[rt<<1|1] += c[rt];
    c[rt<<1] += c[rt];
    c[rt<<1|1] += c[rt];
  }
  c[rt] = 0;
}

void update(int L, int R, int x, int l, int r, int rt)
{
  if (L <= l && r <= R) {
    seg[rt] += x;
    c[rt] += x;
    return ;
  }
  pushdown(rt);
  int mid = (l + r) >> 1;
  if (L <= mid) update(L, R, x, lson);
  if (mid < R) update(L, R, x, rson);
  seg[rt] = max(seg[rt<<1], seg[rt<<1|1]);
}

int query(int x, int l, int r, int rt)
{
  if (l == r) {
    return l;
  }
  pushdown(rt);
  int mid = (l + r) >> 1;
  if (seg[rt<<1] == x) return query(x, lson);
  else return query(x, rson);
}

vector<int> p[N];
int a[N];

int main()
{
  int n, c, k;
  while (read(n)) {
    read(c); read(k);
    for (int i = 1; i <= c; ++ i) {
      p[i].clear();
      p[i].emplace_back(0);
    }
    for (int i = 1; i <= n; ++ i) {
      read(a[i]);
    }
    build(1, n, 1);
    int ans = 0;
    for (int i = 1; i <= n; ++ i) {
      p[a[i]].emplace_back(i);
      int siz = p[a[i]].size();
      update(i, i, c, 1, n, 1);
      update(p[a[i]][siz - 2] + 1, i, -1, 1, n, 1);
      if (siz > k) {
        update(p[a[i]][siz - k - 1] + 1, p[a[i]][siz - k], 1, 1, n, 1);
      }
      if (seg[1] < c) continue;
      int l = query(c, 1, n, 1);
      ans = max(ans, i - l + 1);
    }
    printf("%d\n", ans);
  }
  return 0;
}