C. Codejamon Cipher
Explanation: Google Dynamic Programming Problem (APAC 2017 Round D)
Code:
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#include <bits/stdc++.h> | |
using namespace std; | |
#define INF 1<<30 | |
#define endl '\n' | |
#define maxn 500005 | |
#define tc printf("Case %d: ", cs) | |
#define tcn printf("Case %d:\n", cs); | |
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0); | |
typedef long long ll; | |
const double PI = acos(-1.0); | |
#define dbg(x) cerr << #x << " = " << x << endl; | |
#define dbg2(x, y) cerr << #x << " = " << x << ", " << #y << " = " << y << endl; | |
#define dbg3(x, y, z) cerr << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl; | |
#define dbg4(w,x, y, z) cerr << #w << " = " << w << ", " <<#x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl; | |
const int Mod = 1e9 + 7; | |
map<string, int> cnt; | |
set<string>wordList; | |
int lmt; | |
int len; | |
ll dp[maxn]; | |
string s; | |
ll solve(int pos) | |
{ | |
if (pos == len) return 1; | |
string cur = ""; | |
ll &ans = dp[pos]; | |
if (ans != -1) return ans; | |
ans = 0; | |
int to = pos + lmt + 1; | |
to = min(to, len); | |
for (int i = pos; i < to; i++) { | |
cur += s[i]; | |
sort(cur.begin(), cur.end()); | |
if (cnt[cur]) { | |
ans += (cnt[cur] * solve(i + 1)) % Mod; | |
if (ans >= Mod) ans -= Mod; | |
} | |
} | |
return ans; | |
} | |
int main() | |
{ | |
FASTIO | |
///* | |
//double start_time = clock(); | |
#ifndef ONLINE_JUDGE | |
freopen("in.txt", "r", stdin); | |
freopen("out.txt", "w", stdout); | |
freopen("error.txt", "w", stderr); | |
#endif | |
//*/ | |
int T; | |
cin >> T; | |
for (int cs = 1; cs <= T; cs++) { | |
int n, q; | |
cin >> n >> q; | |
cnt.clear(); | |
wordList.clear(); | |
lmt = 0; | |
for (int i = 0; i < n; i++) { | |
string word; | |
cin >> word; | |
lmt = max(lmt, (int)word.length()); | |
wordList.insert(word); | |
sort(word.begin(), word.end()); | |
cnt[word]++; | |
} | |
cout << "Case #" << cs << ": "; | |
while (q--) { | |
cin >> s; | |
len = s.length(); | |
for (int i = 0; i <= len; i++) dp[i] = -1; | |
cout << solve(0) << " "; | |
} | |
cout << endl; | |
} | |
//double end_time = clock(); | |
//printf( "Time = %lf ms\n", ( (end_time - start_time) / CLOCKS_PER_SEC)*1000); | |
return 0; | |
} |
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