BZOJ 2780: [Spoj]8093 Sevenk Love Oimaster [广义后缀自动机]

JZPGYZ - Sevenk Love Oimaster

    Oimaster and sevenk love each other.

 

    But recently,sevenk heard that a girl named ChuYuXun was dating with oimaster.

As a woman's nature, sevenk felt angry and began to check oimaster's online talk with ChuYuXun.

    Oimaster talked with ChuYuXun n times, and each online talk actually is a string.

Sevenk asks q questions like this,

    "how many strings in oimaster's online talk contain this string as their substrings?"

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这篇文章讲述了一个sevenk如何用后缀自动机检查Oimaster的聊天记录......

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题意：给出很多主串和很多询问串，求一个询问串在主串中出现次数

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裸题啊.....除了我在http://www.cnblogs.com/candy99/p/sam.html中给出的方法

 

还有一种NlogN的做法，求出Parent树DFS序然后就是区间询问多少种不同的颜色，HH的项链.....注意一个点会有多种颜色

 

注意string别用错

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
using namespace std;
const int N=2e5+,M=1e4+;
typedef long long ll;
int n,Q;
string ss[M];
char s[N<<];
struct node{
    int ch[],par,val;
    int cou,cur;
}t[N];
int sz=,root=,last=;
void extend(int c){
    int p=last,np=++sz;
    t[np].val=t[p].val+;
    for(;p&&!t[p].ch[c];p=t[p].par) t[p].ch[c]=np;
    if(!p) t[np].par=root;
    else{
        int q=t[p].ch[c];
        if(t[q].val==t[p].val+) t[np].par=q;
        else{
            int nq=++sz;
            t[nq]=t[q];t[nq].val=t[p].val+;
            t[q].par=t[np].par=nq;
            for(;p&&t[p].ch[c]==q;p=t[p].par) t[p].ch[c]=nq;
        }
    }
    last=np;
}
void solve(){
    int u;
    for(int i=;i<=n;i++){
        u=root; string &s=ss[i];
        for(int j=;j<s.size();j++){
            u=t[u].ch[s[j]-'a'];
            int p=u;
            for(;p&&t[p].cur!=i;p=t[p].par) t[p].cou++,t[p].cur=i;
        }
    }
    while(Q--){
        scanf("%s",s);
        int n=strlen(s),u=root,flag=;
        for(int i=;i<n;i++){
            int c=s[i]-'a';
            if(t[u].ch[c]) u=t[u].ch[c];
            else{flag=;break;}
        }
        if(flag) puts("");
        else printf("%d\n",t[u].cou);
    }
}
int main(){
    freopen("in","r",stdin);
    scanf("%d%d",&n,&Q);
    for(int i=;i<=n;i++){
        scanf("%s",s);
        ss[i]=string(s);
        last=root; int len=ss[i].size();
        for(int j=;j<len;j++) extend(s[j]-'a');
    }
    solve();
}