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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#498986 | #2343. First of Her Name | PetroTarnavskyi | WA | 3ms | 7968kb | C++20 | 2.2kb | 2024-07-30 22:45:16 | 2024-07-30 22:45:16 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define RFOR(i, a, b) for(int i = (a) - 1; i >= (b); i--)
#define SZ(a) int(a.size())
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
#define F first
#define S second
typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef double db;
const int AL = 26;
struct Node
{
int p;
int c;
int g[AL];
int nxt[AL];
int link;
Node(int _c, int _p)
{
c = _c;
p = _p;
fill(g, g + AL, -1);
fill(nxt, nxt + AL, -1);
link = -1;
}
};
struct AC
{
vector<Node> a;
AC(): a(1, {-1, -1}) {}
int addStr(const string& s)
{
int v = 0;
FOR (i, 0, SZ(s))
{
// change to [0 AL)
int c = s[i] - 'a';
if (a[v].nxt[c] == -1)
{
a[v].nxt[c] = SZ(a);
a.PB(Node(c, v));
}
v = a[v].nxt[c];
}
return v;
}
int go(int v, int c)
{
if (a[v].g[c] != -1)
return a[v].g[c];
if (a[v].nxt[c] != -1)
a[v].g[c] = a[v].nxt[c];
else if (v != 0)
a[v].g[c] = go(getLink(v), c);
else
a[v].g[c] = 0;
return a[v].g[c];
}
int getLink(int v)
{
if (a[v].link != -1)
return a[v].link;
if (v == 0 || a[v].p == 0)
return 0;
return a[v].link = go(getLink(a[v].p), a[v].c);
}
};
const int N = 1'000'447;
AC A;
int c[N];
VI g[N];
VI g2[N];
int a[N];
int val[N];
void dfs(int v, int x)
{
x = A.go(x, c[v] - 'A');
val[x]++;
for (auto to : g[v])
dfs(to, x);
}
void dfs2(int v)
{
for (auto to : g2[v])
{
if (to != 0)
{
dfs2(to);
val[v] += val[to];
}
}
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout << fixed << setprecision(15);
int n, k;
cin >> n >> k;
vector<string> v(k);
FOR (i, 0, n)
{
char ch;
int par;
cin >> ch >> par;
c[i + 1] = ch;
g[par].PB(i + 1);
}
FOR (i, 0, k)
{
cin >> v[i];
reverse(ALL(v[i]));
a[i] = A.addStr(v[i]);
}
FOR (i, 0, SZ(A.a))
{
g2[A.getLink(i)].PB(i);
}
dfs(1, 0);
dfs2(0);
FOR (i, 0, k)
{
cout << val[a[i]] << '\n';
}
cerr << double(clock()) / CLOCKS_PER_SEC << '\n';
return 0;
}
詳細信息
Test #1:
score: 0
Wrong Answer
time: 3ms
memory: 7968kb