题目链接:hdu 1227 全文检索
题目大意:略。
解题思路:就是普通的匹配问题,注意相同串,以及输出要按照出现顺序呢。
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 10000 * 60;
const int maxm = 10005;
const int sigma_size = 10;
struct Aho_Corasick {
int sz, g[maxn][sigma_size];
int tag[maxn], fail[maxn], last[maxn];
int vis[maxm], jump[maxm];
int c, ans[maxm];
void init();
int idx(char ch);
void insert(char* str, int k);
void getFail();
void match(char* str);
void put(int u);
void solve(char* s);
}AC;
int N, M;
char s[60005], w[105];
int main () {
while (scanf("%d%d", &M, &N) == 2) {
AC.init();
int mv = 0;
for (int i = 0; i < M; i++) {
scanf("%s", s+mv);
mv += strlen(s+mv);
}
for (int i = 1; i <= N; i++) {
scanf("%*s%*s%*s%s", w);
AC.insert(w, i);
}
AC.solve(s);
}
return 0;
}
void Aho_Corasick::solve(char* s) {
c = 0;
getFail();
match(s);
if (c) {
printf("Found key:");
for (int i = 0; i < c; i++)
printf(" [Key No. %d]", ans[i]);
printf("\n");
} else
printf("No key can be found !\n");
}
void Aho_Corasick::init() {
sz = 1;
tag[0] = 0;
memset(g[0], 0, sizeof(g[0]));
}
int Aho_Corasick::idx(char ch) {
return ch - '0';
}
void Aho_Corasick::put(int u) {
int p = tag[u];
while (p && vis[p] == 0) {
vis[p] = 1;
ans[c++] = p;
p = jump[p];
}
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];
}
jump[k] = tag[u];
tag[u] = k;
}
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]];
}
}
}