2015 Multi-University Training Contest 8 hdu 5384 Danganronpa

Danganronpa

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 318    Accepted Submission(s): 176

Problem Description

给你n个A串，m个B串，对每个A串，询问，这些B串们在该A串中一共出现过多少次

Input

样例个数

n m

接下来n个A串

接下来m个B串

Output

如问题描述，对每个A输出...

Sample Input

1

5 6

orz

sto

kirigiri

danganronpa

ooooo

o

kyouko

dangan

ronpa

ooooo

ooooo

 

 

Sample Output

1

1

0

3

7

 

Source

2015 Multi-University Training Contest 8

 

解题：AC自动机自动AC

 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 250010;
struct trie {
    int word[26],fail,cnt;
    void init() {
        fail = cnt = 0;
        memset(word,-1,sizeof word);
    }
} dic[maxn];
int tot;
char str[1000100];
void insertWord(int root,char *s) {
    for(int i = 0; s[i]; i++) {
        int k = s[i]-'a';
        if(dic[root].word[k] == -1) {
            dic[++tot].init();
            dic[root].word[k] = tot;
        }
        root = dic[root].word[k];
    }
    ++dic[root].cnt;
}
queue<int>q;
void build(int root) {
    while(!q.empty()) q.pop();
    q.push(root);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        for(int i = 0; i < 26; i++) {
            if(dic[u].word[i] == -1) continue;
            if(u == 0) dic[dic[u].word[i]].fail = 0;
            else {
                int v = dic[u].fail;
                while(v && dic[v].word[i] == -1) v = dic[v].fail;
                if(dic[v].word[i] == -1) dic[dic[u].word[i]].fail = 0;
                else dic[dic[u].word[i]].fail = dic[v].word[i];
            }
            q.push(dic[u].word[i]);
        }
    }
}
int query(int root,char *s) {
    int i,ans = 0;
    for(i = 0; s[i]; i++) {
        int k = s[i]-'a';
        while(root && dic[root].word[k] == -1)
            root = dic[root].fail;
        if(dic[root].word[k] != -1) {
            int v = dic[root].word[k];
            while(v) {
                ans += dic[v].cnt;
                v = dic[v].fail;
            }
            root = dic[root].word[k];
        }
    }
    return ans;
}
char FK[100001][10010];
int main() {
    int n,m,t;
    scanf("%d",&t);
    while(t--) {
        dic[tot = 0].init();
        scanf("%d%d",&n,&m);
        for(int i = 0; i < n; ++i)
            scanf("%s",FK[i]);
        while(m--) {
            scanf("%s",str);
            insertWord(0,str);
        }
        build(0);
        for(int i = 0; i < n; ++i)
            printf("%d
",query(0,FK[i]));
    }
    return 0;
}

这才是AC自动机的正确使用方法，Trie图

#include <bits/stdc++.h>
using namespace std;
const int maxn = 250010;
struct Trie{
    int ch[maxn][26],fail[maxn],cnt[maxn],tot;
    int newnode(){
        memset(ch[tot],0,sizeof ch[tot]);
        fail[tot] = cnt[tot] = 0;
        return tot++;
    }
    void init(){
        tot = 0;
        newnode();
    }
    void insert(char *str,int root = 0){
        for(int i = 0; str[i]; ++i){
            if(!ch[root][str[i]-'a'])
                ch[root][str[i]-'a'] = newnode();
            root = ch[root][str[i]-'a'];
        }
        ++cnt[root];
    }
    void build(int root = 0){
        queue<int>q;
        for(int i = 0; i < 26; ++i)
            if(ch[root][i]) q.push(ch[root][i]);
        while(!q.empty()){
            root = q.front();
            q.pop();
            for(int i = 0; i < 26; ++i)
            if(ch[root][i]){
                fail[ch[root][i]] = ch[fail[root]][i];
                cnt[ch[root][i]] += cnt[ch[fail[root]][i]];
                q.push(ch[root][i]);
            }else ch[root][i] = ch[fail[root]][i];
        }
    }
    int query(char *str,int ret = 0,int root = 0){
        for(int i = 0; str[i]; ++i){
            int x = root = ch[root][str[i]-'a'];
            ret += cnt[x];
        }
        return ret;
    }
}ac;
char FK[100001][10010],str[1000010];
int main(){
    int kase,n,m;
    scanf("%d",&kase);
    while(kase--){
        ac.init();
        scanf("%d%d",&n,&m);
        for(int i = 0; i < n; ++i)
            scanf("%s",FK[i]);
        while(m--){
            scanf("%s",str);
            ac.insert(str);
        }
        ac.build();
        for(int i = 0; i < n; ++i)
            printf("%d
",ac.query(FK[i]));
    }
    return 0;
}