CodeForces 1393B Applejack and Storages
A p p l e j a c k Applejack Applejack a n d and and S t o r a g e s Storages Storages
由于本菜鸡很笨，并没有想到什么简单的思路，所以有幸在时间内用权值线段树水过，u1s1我的思维确实垃圾
思路：
利用权值线段树求区间第k小，首先我们只需要知道三个变量（最大相同数量 d m dm dm，次大相同数量 d s ds ds，次次大 d k dk dk)，那么就可以写了，显然这题你用排序肯定不行了，那样最坏时间复杂度是 O ( q l o g ( q + n ) ) O(qlog(q+n)) O(qlog(q+n))，显然会TLE，那么权值线段树可以帮你实现 O ( q l o g ( q + n ) ) O(qlog(q+n)) O(qlog(q+n))，具体思路请看下图：
具体操作看代码：
#include p;
//priority_queue, greater > m;
// ll sum[maxn], a[maxn];
ll max(ll a, ll b) { return a > b ? a : b; }
ll min(ll a, ll b) { return a < b ? a : b; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
map<ll, ll> mp;
ll ksm(ll a, ll b)
{
a %= mod;
ll ans = 1ll;
while (b)
{
if (b & 1)
ans = (ans * a) % mod;
a = (a * a) % mod;
b >>= 1ll;
}
return ans;
}
// ll dp[105][16005];
// string p = "abacaba";
// queue qk, q;
//vector vec;
// ll sumx[maxn], sumy[maxn], sumk[maxn];
int cnt;
map<int, int> p;
struct node
{
int ls, rs, sum;
} tr[maxn * 32];
void inser(int &k, int L, int R, int pos, int w)
{
if (!k)
k = ++cnt;
tr[k].sum += w;
if (L == R)
return;
int mid = L + R >> 1;
if (mid >= pos)
inser(tr[k].ls, L, mid, pos, w);
else
inser(tr[k].rs, mid + 1, R, pos, w);
}
int query(int k, int L, int R, int ik)
{
if (L == R)
return L;
int sk = tr[tr[k].ls].sum;
int mid = L + R >> 1;
if (sk >= ik)
return query(tr[k].ls, L, mid, ik);
else
return query(tr[k].rs, mid + 1, R, ik - sk);
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
//ll t, n, ans;
// scanf("%lld", &t);
// cin >> t;
// while (t--)
// {
p.clear();
int n, k = 0, Q, x, res = 0;
cin >> n;
for (int i = 1; i <= n; i++)
cin >> a[i], p[a[i]]++;
for (int i = 1; i <= n; i++)
{
if (!vis[a[i]])
inser(k, -maxn, maxn, p[a[i]], 1), vis[a[i]] = 1, res++;
}
cin >> Q;
while (Q--)
{
char c;
cin >> c >> x;
if (c == '-')
{
inser(k, -maxn, maxn, p[x], -1), p[x]--;
if (p[x] == 0)
vis[x] = 0, res--;
else
inser(k, -maxn, maxn, p[x], 1);
}
else
{
if (!vis[x])
{
p[x]++, res++, vis[x] = 1;
inser(k, -maxn, maxn, p[x], 1);
}
else
{
inser(k, -maxn, maxn, p[x], -1), p[x]++;
inser(k, -maxn, maxn, p[x], 1);
}
}
//cout << "***" << n << "***" << endl;
int dm = query(k, -maxn, maxn, res);
int ds = query(k, -maxn, maxn, res - 1);
int dk = query(k, -maxn, maxn, res - 2);
//cout << dm << ' ' << ds << ' ' << dk << endl;
if (dm >= 8)
{
cout << "YES" << endl;
continue;
}
if (dm >= 4 && ds >= 4)
{
cout << "YES" << endl;
continue;
}
if (dm >= 4 && ((ds >= 2 && (dm - 4) >= 2) || dk >= 2))
{
cout << "YES" << endl;
continue;
}
cout << "NO" << endl;
}
// }
}