[HDU 2222] Keywords Search AC自动机

http://acm.hdu.edu.cn/showproblem.php?pid=2222

题意：给出 n 个关键词和一个长的字符串，要找出字符串里面出现了多少个关键词。

思路：AC自动机，就是基于字典树的KMP算法，网上博客写的很详细。 还有要注意的就是这题测试数据估计改了，动态分配内存会MLE。只能开数组了。

#include <cstdio>#include <cstring>#include <iostream>using namespace std;/struct node{    int next[26];    int counts, fail;    static int newnode;};int que[250010];node tr[250010];char str[1000010];int node::newnode = 1;//分配没有使用过的下标int Newnode(){    int pos = node::newnode++;    tr[pos].counts = tr[pos].fail =0;    for(int i = 0; i < 26; i++){        tr[pos].next[i] = 0;    }    return pos;}//构造字典树int Build(int rt, char *str){    int i = -1;    int pr = rt;    while(str[++i]){        int pos = str[i]-'a';        if(!tr[pr].next[pos]){            tr[pr].next[pos] = Newnode();        }        pr = tr[pr].next[pos];    }    tr[pr].counts++;  //标记这是一个关键字的结尾    return rt;}void Automaton(int rt)  //AC自动机，利用BFS构造fail{    int rear, top;    rear = top = 0;    que[rear++] = rt;    while(top != rear){        int pr = que[top++];        for(int i = 0; i < 26; i++){            int son = tr[pr].next[i]; //pr 的儿子结点            if(son == NULL)                continue;            if(pr == rt) //如果在根节点的儿子结点匹配失败肯定要跳回根结点                tr[son].fail = pr;            else{                int temp = tr[pr].fail; //通过父结点的fail找子节点的fail                while(temp){                    if(tr[temp].next[i]) //父结点的fail存在第 i 个儿子                    {                        tr[son].fail = tr[temp].next[i]; //指向父结点fail的儿子                        break;                    }                    temp = tr[temp].fail; //继续找上一个fail                }                if(temp == NULL) //不存在就指向rt                    tr[son].fail = rt;            }            que[rear++] = son; //入队        }    }}int query(int rt, char *str){    int pr = rt;    int i = -1, sum = 0;    while(str[++i]){        int pos = str[i] - 'a';        while(!tr[pr].next[pos] && pr != rt) //找匹配成功的串            pr = tr[pr].fail;        pr = (tr[pr].next[pos] == NULL) ? rt : tr[pr].next[pos]; //成功或者失败        int temp = pr;        while(temp != rt && tr[temp].counts != -1) //成功就累加        {            sum += tr[temp].counts;            tr[temp].counts = -1; //标记被计算过了            temp = tr[temp].fail;        }    }    return sum;}int main(){    int Test;    scanf("%d", &Test);    while(Test--){        int n;        scanf("%d", &n);        node::newnode = 1;        int rt = Newnode();        for(int i = 0; i < n; i++){            scanf("%s", str);            rt = Build(rt, str);        }        Automaton(rt);        scanf("%s", str);        printf("%d\n", query(rt, str));    }    return 0;}