题目连接:hdu 3695 Computer Virus on Planet Pandora
题目大意:给定一些病毒串,要求推断说给定串中包括几个病毒串,包括反转。
解题思路:将给定的字符串展开,然后匹配一次后反转后再匹配一次。
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 250000;
const int sigma_size = 26;
struct Aho_Corasick {
int sz, g[maxn][sigma_size];
int tag[maxn], fail[maxn], last[maxn];
int vis[300], jump[300];
void init();
int idx(char ch);
void insert(char* str, int k);
void getFail();
void match(char* str);
void put(int u);
int solve(char* s);
}AC;
int N;
char w[1005], s[5100005];
int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
AC.init();
scanf("%d", &N);
for (int i = 1; i <= N; i++) {
scanf("%s", w);
AC.insert(w, i);
}
int mv = 0, x;
char ch;
getchar();
while(ch = getchar(), ch != '
') {
if (ch == '[') {
scanf("%d%c", &x, &ch);
for (int i = 0; i < x; i++)
s[mv++] = ch;
getchar();
} else
s[mv++] = ch;
}
s[mv] = ' ';
printf("%d
", AC.solve(s));
}
return 0;
}
int Aho_Corasick::solve(char* s) {
memset(vis, 0, sizeof(vis));
AC.getFail();
AC.match(s);
reverse(s, s + strlen(s));
AC.match(s);
int ans = 0;
for (int i = 1; i <= N; i++)
if (vis[jump[i]])
ans++;
return ans;
}
void Aho_Corasick::init() {
sz = 1;
tag[0] = 0;
memset(g[0], 0, sizeof(g[0]));
}
int Aho_Corasick::idx(char ch) {
return ch - 'A';
}
void Aho_Corasick::put(int u) {
if (vis[tag[u]])
return;
vis[tag[u]] = 1;
if (last[u])
put(last[u]);
}
void Aho_Corasick::insert(char* str, int k) {
int u = 0, n = strlen(str);
for (int i = 0; i < n; i++) {
int v = idx(str[i]);
if (g[u][v] == 0) {
tag[sz] = 0;
memset(g[sz], 0, sizeof(g[sz]));
g[u][v] = sz++;
}
u = g[u][v];
}
if (tag[u] == 0)
tag[u] = k;
jump[k] = tag[u];
}
void Aho_Corasick::match(char* str) {
int n = strlen(str), u = 0;
for (int i = 0; i < n; i++) {
int v = idx(str[i]);
while (u && g[u][v] == 0)
u = fail[u];
u = g[u][v];
if (tag[u])
put(u);
else if (last[u])
put(last[u]);
}
}
void Aho_Corasick::getFail() {
queue<int> que;
for (int i = 0; i < sigma_size; i++) {
int u = g[0][i];
if (u) {
fail[u] = last[u] = 0;
que.push(u);
}
}
while (!que.empty()) {
int r = que.front();
que.pop();
for (int i = 0; i < sigma_size; i++) {
int u = g[r][i];
if (u == 0) {
g[r][i] = g[fail[r]][i];
continue;
}
que.push(u);
int v = fail[r];
while (v && g[v][i] == 0)
v = fail[v];
fail[u] = g[v][i];
last[u] = tag[fail[u]] ? fail[u] : last[fail[u]];
}
}
}