【字符串·ac自动机】uva1449Dominating Patterns

ac自动机的基本用途是实现多个模板串在一个文本串中的匹配
如：a acd bbca aca 在 aacdacacabbcac中分别出现了几次？

在trie树上计算一些新的东西；
失配数组f[maxn]: 如果在文本串I点失配，那还能去哪个节点碰碰运气
后缀链接last[maxn]: 节点j沿着失配链接往回走，遇到的下一个单词节点编号；

题目uva4670，找出出现次数最多的字符串，用不同的数表示不同的单词结尾，将print函数改为字符串计数器+1；

模板串可能有重复，后一个字符串的val会覆盖前一个字符串，这时用一个映射，两个字符串可以映射到同一个索引，答案输出两遍；

#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<queue>#include<map>using namespace std;typedef long long ll;const int maxn = 1000010;int ch[maxn][26];int val[maxn];int sz;int idx(char c){ return c - 'a'; }char P[155][75]; //数组要开大一点map<string, int> ms;char str[maxn];void insert(char *s, int v){    int u = 0, n = strlen(s);    for (int i = 0; i < n; i++){        int c = idx(s[i]);        if (!ch[u][c]){            memset(ch[sz], 0, sizeof(ch[sz]));            val[sz] = 0;            ch[u][c] = sz++;        }        u = ch[u][c];    }    val[u] = v;    ms[string(s)] = v;}int n;int nex[maxn];int last[maxn];int cnt[maxn];void init(){    ms.clear();    sz = 1;    memset(ch[0], 0, sizeof(ch[0]));    memset(cnt, 0, sizeof(cnt));    for (int i = 1; i <= n; i++){        scanf("%s", P[i]);        insert(P[i], i);    }}void getNext(){    memset(nex, 0, sizeof(nex));    memset(last, 0, sizeof(last));    queue<int> q;    nex[0] = 0;    for (int c = 0; c < 26; c++){        int u = ch[0][c];        if (u) { nex[u] = 0; q.push(u); last[u] = 0; }    }    while (!q.empty()){        int r = q.front(); q.pop();        for (int c = 0; c < 26; c++){            int u = ch[r][c];            if (!u) continue;            q.push(u);            int v = nex[r];            while (v&&!ch[v][c]) v = nex[v];            nex[u] = ch[v][c];            last[u] = val[nex[u]] ? nex[u] : last[nex[u]];        }    }}void print(int j){    if (j){        cnt[val[j]]++;        print(last[j]);    }}void find(char *t){    int len = strlen(t);    int j = 0;    for (int i = 0; i < len; i++){        int c = idx(t[i]);        while (j&&!ch[j][c]) j = nex[j];        j = ch[j][c];        if (val[j]) print(j);        else if (last[j]) print(last[j]);    }}int main(){    while (~scanf("%d", &n) && n){        init();        getNext();        scanf("%s", str);        find(str);        int top = -1;        for (int i = 1; i <= n; i++){            if (cnt[i] > top) top = cnt[i];        }        printf("%d\n", top);        for (int i = 1; i <= n; i++){            if (cnt[ms[string(P[i])]] == top) printf("%s\n", P[i]);        }    }    return 0;}