题目链接

Codeforces Round 977 (Div. 2) C2 Adjust The Presentation (Hard Version)

思路

我们令$s[i]$表示第$i$名成员最早上场的时间。

显然，只有当$s$数组单调不减时才会有解。

换句话说，只有$s[i]\le s[i+1]$时才有解。

因此，我们可以使用$set$来动态维护有多少个$s[i]\le s[i+1]$，当$set$的大小为$0$时有解，否则无解。

代码

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 2e5 + 5;
const int mod = 998244353;
const int inf = 1e6;
int n, m, q;
int a[N], b[N], s[N], idx[N];
void solve()
{cin >> n >> m >> q;for (int i = 1; i <= n; i++){cin >> a[i];idx[a[i]] = i;}map<int, set<int>> mp;for (int i = 1; i <= m; i++){cin >> b[i];mp[b[i]].insert(i);}for (int i = 1; i <= n; i++){if (!mp.count(a[i])){mp[a[i]].insert(inf);}s[i] = *mp[a[i]].begin();}s[n + 1] = inf;set<int> st;for (int i = 1; i <= n; i++){if (s[i] > s[i + 1]){st.insert(i);}}if (!st.size()){cout << "YA" << endl;}elsecout << "TIDAK" << endl;while (q--){int id, t;cin >> id >> t;int res = idx[b[id]];mp[b[id]].erase(mp[b[id]].find(id));if (mp[b[id]].size())s[res] = *(mp[b[id]].begin());elses[res] = inf;if (st.count(res))st.erase(res);if (res > 1 && st.count(res - 1))st.erase(res - 1);if (s[res] > s[res + 1])st.insert(res);if (s[res - 1] > s[res] && res > 1)st.insert(res - 1);b[id] = t;res = idx[t];mp[t].insert(id);s[res] = *mp[t].begin();if (st.count(res))st.erase(res);if (res > 1 && st.count(res - 1))st.erase(res - 1);if (s[res] > s[res + 1])st.insert(res);if (s[res - 1] > s[res] && res > 1)st.insert(res - 1);if (!st.size()){cout << "YA" << endl;}elsecout << "TIDAK" << endl;}
}
signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int test = 1;cin >> test;for (int i = 1; i <= test; i++){solve();}return 0;
}