# # Educational Codeforces Round 97 (CF1437) 题解

## # Problem A - Marketing Scheme (opens new window)

def read_int():
return int(input())

return map(int, input().split(' '))

for case_num in range(t):
print('YES' if l * 2 > r else 'NO')

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## # Problem B - Reverse Binary Strings (opens new window)

def read_int():
return int(input())

return map(int, input().split(' '))

for case_num in range(t):
s = input()
tot = 0
for i in range(n - 1):
if s[i] == s[i + 1]:
tot += 1
print((tot + 1) // 2)

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## # Problem C - Chef Monocarp (opens new window)

1. 我们不会用到超过$2n$的时间。
2. 数组的原始顺序没有影响。

（Python 3）
def read_int():
return int(input())

return map(int, input().split(' '))

inf = int(1e9)
for case_num in range(t):
a.sort()
dp = [0 for _ in range(n * 2 + 1)]
for t in a:
ndp = [inf for _ in range(n * 2 + 1)]
lo = inf
for i in range(n * 2):
lo = min(lo, dp[i])
ndp[i + 1] = min(ndp[i + 1], lo + abs(i + 1 - t))
dp = ndp
print(min(dp))

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## # Problem D - Minimal Height Tree (opens new window)

#include <cstdio>
#include <iostream>
#include <vector>

using namespace std;

template <typename T> void read(T &x) {
x = 0;
char c = getchar();
T sig = 1;
for (; !isdigit(c); c = getchar())
if (c == '-')
sig = -1;
for (; isdigit(c); c = getchar())
x = (x << 3) + (x << 1) + c - '0';
x *= sig;
}

class Solution {
public:
void solve() {
int n;
int height = 0;
vector<int> level = {1};
int col = 0, first, last = 0;
for (int i = 0; i < n - 1; ++i) {
int u;
if (u < last) {
col++;
if (col == level[height]) {
height++;
col = 0;
}
}
if (level.size() <= height + 1)
level.emplace_back(0);
level[height + 1]++;
last = u;
}
printf("%d\n", (int)level.size() - 1);
}
};

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int t;
while (t--) {
Solution solution = Solution();
solution.solve();
}
}
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## # Problem E - Make It Increasing (opens new window)

1. 我们可以在数组$a$$b$的头尾各添加一个哨兵位，这样就可以把原问题分解为若干子问题。
2. 每个子问题实质上是最长上升子序列（LIS）。

• $l>r$，也即空区间的情形，对应的结果显然为$0$
• $l=r$，也即区间长度为$1$的情形，对应的结果取决于$a[l]$是否在$[lo,hi]$范围内。
• $l。在这一情形中，我们需要找到区间内的最长上升子序列。不过，这与经典的LIS问题有两点区别。首先，如果当前数字不在其合法范围内（这一范围可以由$lo$$hi$计算得到，为$[lo+i-l,hi-r+i]$），它就不能被加入到LIS中。另一区别是，我们不能直接使用$a[i]$。考虑这一例子，$[2,5,3,1]$。我们可以用$[2,3]$作为LIS吗？答案是否定的，因为如果这样的话，中间的数字就无法安排了。我们如何将这一距离因素考虑进来呢？方法是，用$a[i]-i$代替$a[i]$，同时，求取最长不下降子序列而非最长上升子序列。通过$-i$，我们就消去了不同位置间的距离差异。

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <vector>
#define INF 0x3f3f3f3f

using namespace std;

template <typename T> void read(T &x) {
x = 0;
char c = getchar();
T sig = 1;
for (; !isdigit(c); c = getchar())
if (c == '-')
sig = -1;
for (; isdigit(c); c = getchar())
x = (x << 3) + (x << 1) + c - '0';
x *= sig;
}

class Solution {
public:
void solve() {
int n, k;
vector<int> a(n + 2), b(k + 2);
a[0] = -INF, a[n + 1] = INF;
b[0] = 0, b[k + 1] = n + 1;
for (int i = 1; i <= n; ++i)
for (int i = 1; i <= k; ++i) {
if (a[b[i]] - a[b[i - 1]] < b[i] - b[i - 1]) {
printf("-1");
return;
}
}
int ans = 0;
auto handle = [&](int l, int r, int lo, int hi) {
if (l > r)
return 0;
if (l == r)
return (a[l] >= lo && a[l] <= hi) ? 0 : 1;
vector<int> LIS;
for (int i = l; i <= r; ++i) {
int clo = lo + i - l, chi = hi - r + i;
if (a[i] < clo || a[i] > chi)
continue;
int pos = upper_bound(LIS.begin(), LIS.end(), a[i] - i) - LIS.begin();
if (pos >= LIS.size())
LIS.emplace_back(a[i] - i);
else
LIS[pos] = a[i] - i;
}
return r - l + 1 - (int)LIS.size();
};
for (int i = 1; i <= k + 1; ++i)
ans += handle(b[i - 1] + 1, b[i] - 1, a[b[i - 1]] + 1, a[b[i]] - 1);
printf("%d", ans);
}
};

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
Solution solution = Solution();
solution.solve();
}
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## # Problem F - Emotional Fishermen (opens new window)

1. $dp[i]$$dp[i]$。这一转移要求我们从$[0,pre[i]]$中选择一个元素加入排列。我们可以通过当前集合中的元素个数来求出可能的选择数。
2. $dp[j]$$dp[i]$，其中$j\leq pre[i]$。为了加速计算这一类转移，我们可以在每次循环后计算$dp$数组的前缀和。

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <vector>
#define MOD 998244353

using namespace std;
typedef long long ll;

template <typename T> void read(T &x) {
x = 0;
char c = getchar();
T sig = 1;
for (; !isdigit(c); c = getchar())
if (c == '-')
sig = -1;
for (; isdigit(c); c = getchar())
x = (x << 3) + (x << 1) + c - '0';
x *= sig;
}

class Solution {
public:
void solve() {
int n;
vector<int> a(n);
for (int i = 0; i < n; ++i)
sort(a.begin(), a.end());
if (a[n - 2] * 2 > a[n - 1]) {
printf("0");
return;
}

vector<int> pre(n, -1);
int l = n - 2;
for (int r = n - 1; r >= 0; --r) {
while (l >= 0 && a[l] * 2 > a[r])
l--;
if (l >= 0)
pre[r] = l;
}

ll ans = 0;
vector<ll> dp(n, 1), S(n + 1);
for (int i = 1; i <= n; ++i)
S[i] = i;
for (int i = 1; i < n; ++i) {
vector<ll> ndp(n, 0);
for (int j = 0; j < n; ++j) {
// Case 1: j to j
int left = pre[j] == -1 ? 0 : pre[j] + 1;
left -= i - 1;
if (left > 0)
ndp[j] = (ndp[j] + dp[j] * left % MOD);

// Case 2: smaller to j
if (pre[j] != -1)
ndp[j] = (ndp[j] + S[pre[j] + 1]) % MOD;
}
for (int j = 1; j <= n; ++j)
S[j] = (S[j - 1] + ndp[j - 1]) % MOD;
dp = move(ndp);
}
printf("%lld", S[n]);
}
};

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
Solution solution = Solution();
solution.solve();
}
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## # Problem G - Death DBMS (opens new window)

#include <iostream>
#include <queue>
#include <set>
#include <unordered_map>
#include <vector>

using namespace std;
struct Node {
int idx = -1, fail = 0, children[26]{};
};

int main() {
int n, q;
cin >> n >> q;

vector<string> names(n + 1);
unordered_map<string, int> dict;
unordered_map<int, int> id_dict;
vector<set<pair<int, int>, greater<>>> heaps;
vector<int> suspicion(n + 1);
suspicion[0] = -1;
vector<Node> nodes = {Node{}};
int idx = 0;
for (int i = 1; i <= n; ++i) {
cin >> names[i];
if (dict.count(names[i])) {
heaps[dict[names[i]]].emplace(0, i);
id_dict[i] = dict[names[i]];
continue;
}
dict[names[i]] = idx;
id_dict[i] = idx;
heaps.push_back({{0, i}});
int p = 0;
for (char c : names[i]) {
if (!nodes[p].children[c - 'a']) {
nodes[p].children[c - 'a'] = nodes.size();
nodes.emplace_back(Node{});
}
p = nodes[p].children[c - 'a'];
}
nodes[p].idx = idx++;
}

queue<int> que, visited;
vector<bool> vis(nodes.size());
for (int u : nodes[0].children)
if (u)
que.push(u);
while (!que.empty()) {
int u = que.front();
que.pop();
for (int i = 0; i < 26; ++i) {
auto &v = nodes[u].children[i];
if (v) {
nodes[v].fail = nodes[nodes[u].fail].children[i];
que.push(v);
} else
v = nodes[nodes[u].fail].children[i];
}
}

string output;

auto query = [&](string &s) {
int ans = -1;
int p = 0;
int idx = 0;
while (idx < s.size()) {
char c = s[idx];
if (nodes[p].children[c - 'a']) {
p = nodes[p].children[c - 'a'];
if (!vis[p]) {
vis[p] = true;
que.push(p);
}
idx++;
} else {
p = nodes[p].fail;
if (!p)
idx++;
}
}
while (!que.empty()) {
int u = que.front();
que.pop();
visited.push(u);
if (nodes[u].idx != -1)
ans = max(ans, heaps[nodes[u].idx].begin()->first);
if (nodes[u].fail && !vis[nodes[u].fail]) {
vis[nodes[u].fail] = true;
que.push(nodes[u].fail);
}
}
while (!visited.empty()) {
vis[visited.front()] = false;
visited.pop();
}
output += to_string(ans) + "\n";
};

while (q--) {
int t;
cin >> t;
if (t == 1) {
int i, x;
cin >> i >> x;
heaps[id_dict[i]].erase({suspicion[i], i});
suspicion[i] = x;
heaps[id_dict[i]].emplace(suspicion[i], i);
} else {
string s;
cin >> s;
query(s);
}
}

cout << output;
}
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