QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#504723 | #9111. Zayin and String | WorldFinalEscaped# | TL | 0ms | 0kb | C++14 | 5.2kb | 2024-08-04 15:23:52 | 2024-08-04 15:23:52 |
answer
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 4005;
const int M = 4005;
const double eps = 1e-12;
const double inf = 1e15;
const int mod = 998244353;
inline int qpow(int a, int b = mod - 2) {
int res = 1;
while (b) {
if (b & 1) res = 1ll * res * a % mod;
a = 1ll * a * a % mod;
b >>= 1;
}
return res;
}
char a[N], b[M];
int n, K;
vector<int> adj[M];
int fail[M];
int ch[M][26], actot;
ll val[M];
void dfs0(int u) {
for (auto v: adj[u]) {
val[v] += val[u];
dfs0(v);
}
}
void ac_build() {
queue<int> q;
for (int i = 0; i < 26; i++)
if (ch[0][i])
q.push(ch[0][i]), fail[ch[0][i]] = 0;
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = 0; i < 26; i++) {
int &go = ch[u][i];
if (go) {
fail[go] = ch[fail[u]][i];
q.push(go);
} else {
go = ch[fail[u]][i];
}
}
}
for (int i = 1; i <= actot; i++) {
adj[fail[i]].push_back(i);
}
dfs0(0);
}
// f[i][j]: a[1...i], ac node j
double f[N][M], g[N][M];
inline void ckmax(double &x, double y) {
if (x < y) x = y;
}
bool check(double ANS) {
f[0][0] = 0, g[0][0] = -inf;
for (int j = 1; j <= actot; j++) {
f[0][j] = g[0][j] = -inf;
}
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= actot; j++) {
f[i][j] = -inf, g[i][j] = -inf;
}
for (int j = 0; j <= actot; j++) {
// not choose a[i]
ckmax(f[i][j], f[i - 1][j]);
ckmax(g[i][j], g[i - 1][j]);
// choose a[i]
int trans = ch[j][a[i] - 'a'];
ckmax(g[i][trans], max(f[i - 1][j], g[i - 1][j]) + val[trans] - ANS);
}
// for (int j = 0; j <= actot; j++) {
// printf("f[%d][%d] = %.10f\n", i, j, f[i][j]);
// printf("g[%d][%d] = %.10f\n", i, j, g[i][j]);
// }
}
double t = *max_element(g[n], g[n] + actot + 1);
return t >= -eps;
}
pair<int, int> where[N][M];
void check2(double ANS) {
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= actot; j++) {
where[i][j] = {-1, -1};
}
}
f[0][0] = 0, g[0][0] = -inf;
for (int j = 1; j <= actot; j++) {
f[0][j] = g[0][j] = -inf;
}
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= actot; j++) {
f[i][j] = -inf, g[i][j] = -inf;
}
for (int j = 0; j <= actot; j++) {
// not choose a[i]
ckmax(f[i][j], f[i - 1][j]);
if (g[i][j] < g[i - 1][j])
g[i][j] = g[i - 1][j], where[i][j] = {i - 1, j};
// choose a[i]
int trans = ch[j][a[i] - 'a'];
if (g[i][trans] < f[i - 1][j] + val[trans] - ANS)
g[i][trans] = f[i - 1][j] + val[trans] - ANS, where[i][trans] = {i - 1, -1};
if (g[i][trans] < g[i - 1][j] + val[trans] - ANS)
g[i][trans] = g[i - 1][j] + val[trans] - ANS, where[i][trans] = {i - 1, j};
}
}
int I = n, J = max_element(g[n], g[n] + actot + 1) - g[n];
vector<int> pos;
while (J != -1) {
if (where[I][J].second == -1) {
pos.push_back(I);
break;
}
if (where[I][J].second != J)
pos.push_back(I);
int i = where[I][J].first, j = where[I][J].second;
I = i, J = j;
}
reverse(pos.begin(), pos.end());
// cout << "ways = ";
// for (auto it: pos) {
// cout << it << ' ';
// }
// cout << '\n';
int cur = 0;
ll vals = 0;
for (auto p: pos) {
cur = ch[cur][a[p] - 'a'];
vals += val[cur];
}
// printf("vals = %lld\n", vals);
printf("%lld\n", vals % mod * qpow(pos.size()) % mod);
}
void sc() {
while (actot >= 0) {
memset(ch[actot], 0, sizeof(ch[actot]));
fail[actot] = 0;
val[actot] = 0;
adj[actot].clear();
actot--;
}
actot = 0;
scanf("%d%d", &n, &K);
scanf("%s", a + 1);
while (K--) {
scanf("%s", b + 1);
int cur = 0;
for (int i = 1; b[i] != '\0'; i++) {
int &go = ch[cur][b[i] - 'a'];
if (!go) go = ++actot;
cur = go;
}
int _; scanf("%d", &_);
val[cur] += _;
}
ac_build();
// for (double t = -10; t <= 10; t += 1.0) {
// printf("check(%d) = %d\n", t, check(t));
// }
double L = -2e8, R = 2e8;
while (fabs(R - L) > eps) {
double MID;
if (fabs(R - L) > 1) MID = (L + R) / 2.0;
else MID = sqrt(L * R);
if (check(MID)) L = MID;
else R = MID;
}
// printf("%.10f\n", L);
check2(L);
}
int main() {
int T; scanf("%d", &T);
while (T--) sc();
return 0;
}
/*
3
17 3
woyaoakccpcxiamen
ak 5
ccpc 6
xiamen 8
33 3
niweishenmezhengtianxunlianbuliwo
wo 3
se 1
zayin 7
39 2
programmingcontestandmewhichisimportant
me 11
gg 2
*/
/*
1
5 3
aaaaa
a 5
aa 4
aaaaa 100
*/
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 0
Time Limit Exceeded
input:
80 4 7 ggeg g 92946 d 65678 gg 50828 wrj 93954 gge 53780 a 94179 geg 40837 5 6 hiiii ii 67984 foh 69185 hi 88153 pj 39431 i 32219 wfmve 96834 8 12 wvxxvwww xw 1522 rzl 16262 wx 77596 v 69622 vw 82225 nykkmkv 19236 xv 83470 o 16781 w 2405 m 98319 ww 13889 qggbvty 16331 8 14 jjjiiijh i 96670 z 74397 i...