掃黃機器人
謝一
製作動機
黃
實體掃黃
線上掃黃
掃黃機制
建立字典
字串匹配
HASH
\(Hash(S) = S_0 * r^0 + S_1 * r^1 + \dots + S_n * r^n \ mod P\)
檢查 hash 值是否一樣就能判斷兩個字串是否相同
重疊問題？
檢查輸出字串中包含黃時：
扣
import discord
r, mod = 283, 2147483647;
T = open("token.txt", 'r').read();
keyword = open("keyword.txt", 'r').read().split();
KW = [];
ints = discord.Intents.all();
bot = discord.Client(intents = ints);
def cut(S : str):
T = "";
for s in S:
if ord(s) < 128 and (ord(s) < ord('a') or ord(s) > ord('z')):
continue;
else:
T += s;
return T;
def has(S : str):
E = [];
h, j, k, x = 0, 0, 0, 1;
for i in range(len(S)):
if S[i] >= 'a' and S[i] <= 'z':
E.append(j);
j += 1;
else:
j += 3;
j = 0;
S = S.encode("utf-8");
for i in range(len(S)):
if k < len(E) and i == E[k]:
h = (h + (S[i] + 256) * x) % mod;
k += 1;
else:
h = (h + S[i] * x) % mod;
x = x * r % mod;
j += 1;
return [h, j];
def hsh(S : str):
E, H = [], [0];
j, k, x = 0, 0, 1;
for i in range(len(S)):
if S[i] >= 'a' and S[i] <= 'z':
E.append(j);
j += 1;
else:
j += 3;
j = 0;
S = S.encode("utf-8");
for i in range(len(S)):
if k < len(E) and i == E[k]:
H.append((H[i] + (S[i] + 256) * x) % mod);
k += 1;
else:
H.append(((H[i] + S[i] * x)) % mod);
x = x * r % mod;
return H;
def yellow(msg : str):
S = hsh(msg);
for [k, w] in KW:
for i in range(1, len(S) - w + 1):
if k == (S[i + w - 1] - S[i - 1] + mod) % mod:
return 1;
k = k * r % mod;
return 0;
@bot.event
async def on_ready():
for kw in keyword:
KW.append(has(kw));
print("logged in as", end = ' ');
print(bot.user);
@bot.event
async def on_message(msg : discord.Message):
if msg.author == bot.user:
return;
M = msg.content;
M = cut(M.lower());
if yellow(M):
print("業績 + 1");
await msg.channel.send(file = discord.File("swipe.png"));
await msg.channel.send(file = discord.File("yellow.png"));
if __name__ == "__main__":
bot.run(T);實測時間
掃黃！！！
DigoliangNeverDie
Subtask 1: \(N \le 100\)
Enumerate \(l, r\) and find the minimum \(H_i \ (l \le i \le r)\), then compare all \(H_i \times (r - l + 1)\).
Time complexity: enumerate \(l, r \ \times\) find min = \(\mathcal{O}(N^2) \times \mathcal{O}(N) = \mathcal{O}(N^3)\)
Subtask 2: \(N \le 1919\)
Notice That:
For the same \(r\), we can decrease \(l\) from \(r\) to \(1\) and update the minimum \(H_i\) between \(l\) and \(r\) at the same time, which makes the time complexity drop to \(\mathcal{O}(N^2)\)
Subtask 3
For all \(i\), we only have to find the greatest \(l \ \text{s.t.} \ l < i \cap H_l < H_i\) and the smallest \(r \ \text{s.t.} \ i < r \cap H_r < H_i\), then the interval that height \(H_i\) can cover is \((l, r)\).
This can be done by a monotone stack in amortized \(\mathcal{O}(1)\) time.
Monotone Stack
push(\(H_i\)) : check if \(H_i\) is smaller than or equal to the top element, if true then the top element will never be used (since it is larger than or equal to \(H_i\) and has a smaller index than \(i\) so it will not be a \(l\) for any \(j > i\)), so we can pop it out and recursively check the top element until \(H_i\) is greater than the top element, which is also when \(H_i\) meets its \(l\).
Solution
For each \(i\), we can find its \(l\) in \(\mathcal{O}(N)\) time by monotone stack. To find \(r\), we only have to reverse to array and run again the same process.
Thus, the overall time complexity for this solution is \(\mathcal{O}(N)\)
#include <bits/stdc++.h>
#define int long long
#define ff first
#define ss second
using namespace std;
int H[11451481], L[11451481], R[11451481];
signed main(){
cin.tie(0), cout.tie(0), ios::sync_with_stdio(0);
int n, ans = 0;
cin >> n;
stack<pair<int, int>> S;
S.push({-1, 0});
for(int i = 1; i <= n; i++){
cin >> H[i];
while(H[i] <= S.top().ff) S.pop();
L[i] = S.top().ss, S.push({H[i], i});
}
while(!S.empty()) S.pop();
S.push({-1, n + 1});
for(int i = n; i; i--){
while(H[i] <= S.top().ff) S.pop();
R[i] = S.top().ss, S.push({H[i], i});
}
for(int i = 1; i <= n; i++) ans = max(ans, (R[i] - L[i] - 1) * H[i]);
cout << ans << "\n";
return 0;
}掃黃機器人
By thanksone
