题目大意:略。
解题思路:先建立AC自动机,然后在AC自动机中开一个数组,用来记录说每个不读特征码出现的次数。
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 50005;
const int sigma_size = 256;
struct Aho_Corasick {
int sz, g[maxn][sigma_size];
int tag[maxn], fail[maxn], last[maxn];
int cnt[1005];
void init();
int idx(char ch);
void insert(char* str, int k);
void getFail();
void match(char* str);
void put(int u);
}A;
int N;
char s[1005][55], str[2000005];
int main () {
while (scanf("%d", &N) == 1) {
A.init();
for (int i = 1; i <= N; i++) {
scanf("%s", s[i]);
A.insert(s[i], i);
}
A.getFail();
scanf("%s", str);
A.match(str);
for (int i = 1; i <= N; i++)
if (A.cnt[i])
printf("%s: %d\n", s[i], A.cnt[i]);
}
return 0;
}
void Aho_Corasick::init() {
sz = 1;
tag[0] = 0;
memset(g[0], 0, sizeof(g[0]));
}
int Aho_Corasick::idx(char ch) {
return ch;
}
void Aho_Corasick::put(int u) {
cnt[tag[u]]++;
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];
}
tag[u] = k;
}
void Aho_Corasick::match(char* str) {
memset(cnt, 0, sizeof(cnt));
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]];
}
}
}