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!!!
DigoliangSchrödingerAlive
Def: \(u \xrightarrow{k} v\) = the count of different paths starting from \(u\) ending at \(v\) through exactly \(k\) edges
Let \(G\) be the original adjacency matrix, then \(G^n[i][j]\) = \(i \xrightarrow{n} j\)
\(G^n[i][j] = \sum_{k = 1}^{N}{G^{(n - 1)}[i][k] \times G[k][j]}\)
\(u \xrightarrow{k + 1} v = \sum_{w \in V}^{}(u \xrightarrow{k} w)(w \xrightarrow{1} v)\)
Time Complexity
Time = Matrix multiplication \(\times\) multiplication count = \(\mathcal{O}(N^3) \times \mathcal{O}(K) = \mathcal{O}(N^3K)\)
Since the multiplication of matrices have associative property, so \(G^n = (G^{\frac{n}{2}})^2\)
\(\implies\) multilication count \(\mathcal{O}(K) \to \mathcal{O}(\lg{K})\)
\(\implies\) time complexity = \(\mathcal{O}(N^3\lg{K})\)
Speed up
Code
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int mod = 1919810513;
array<array<int, 120>, 120> G;
array<array<int, 120>, 120> mul(int n, array<array<int, 120>, 120> &A, array<array<int, 120>, 120> &B){
array<array<int, 120>, 120> C;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
C[i][j] = 0;
for(int k = 1; k <= n; k++) C[i][j] += A[i][k] * B[k][j] % mod;
C[i][j] %= mod;
}
}
return C;
}
array<array<int, 120>, 120> hayahayahatimi(int n, int k){
array<array<int, 120>, 120> P;
for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) P[i][j] = (i == j);
for(int i = 1; i <= k; i <<= 1){
if(i & k) P = mul(n, P, G);
G = mul(n, G, G);
}
return P;
}
signed main(){
int n, m, k, u, v;
cin >> n >> m >> k;
while(m--){
cin >> u >> v;
G[u][v] = 1;
}
array<array<int, 120>, 120> P = hayahayahatimi(n, k);
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++) cout << P[i][j] << " \n"[j == n];
}
return 0;
}
掃黃機器人
By thanksone
掃黃機器人
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