HDU - 6096 处理后缀的字典树

题意:给定n个字符串,m次询问,每次询问多少个字符串前缀是pre且后缀是suf,前后缀不可相交

字典树同时存储前后缀,假设字符串长为len则更新2*len个节点,依次按s[0],s[len-1],s[1],s[len-2],s[2]...更新

对于询问,按pre[0],suf[len2-1],pre[1],suf[len2-2],...pre[len1-1],suf[0]查询

如果前后缀不等长则用特殊字符填充,询问若碰到特殊字符就当前层暴搜(由于后一层前后缀肯定存在确定字符,所以不会被卡)

必须要确保的是询问时字典树中任一字符串必须长度大于等于当前查询的前后缀长度和(xxx xx xx非法)

因此需要离线处理

时间复杂度\(O(能过)\)

#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define println(a) printf("%lld\n",(ll)a)
using namespace std;
const int MAXN = 2e5 + 11;
const int MOD = 1e9+7;
typedef long long ll;
const ll INF = 1ll<<62;
ll read(){
    ll x = 0, f = 1; register char ch = getchar();
    while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
    while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x*f;
}
string str[MAXN];
bool cmp(const string &a,const string &b){
    return a.length()<b.length();
}
struct Q{
    string pre,suf;
    int id,len;
    Q(){};
    Q(string a,string b,int _id){
        pre=a;
        suf=b;
        id=_id;
        len=pre.length();
        len+=suf.length();
        if(pre.length()<suf.length()){
            for(int i=pre.length();i<suf.length();i++){
                pre.push_back('@');
            }
        }else if(suf.length()<pre.length()){
            string t((int)pre.length()-suf.length(),'@');
            suf=t+suf;
        }
    }
    bool operator < (const Q &QAQ) const{
        return len<QAQ.len;
    }
}q[MAXN];
int ans[MAXN];
const int MAXV = 3e6+11;
struct Trie{
    int son[MAXV][26];
    int size[MAXV];
    int tot,root;
    void init(){
        root=tot=1;
        size[root]=0;
        rep(i,0,25) son[root][i]=0;
    }
    int node(){
        ++tot;
        size[tot]=0;
        rep(i,0,25) son[tot][i]=0;
        return tot;
    }
    void insert(const string &str){
        int o=root,t;
        int cur1=0,cur2=str.length();
        int len=2*str.length();
        for(int i=0;i<len;i++){
            if(i&1) t=--cur2;
            else t=cur1++;
            size[o]++;
            int wh=str[t]-'a';
            if(son[o][wh]==0) son[o][wh]=node();
            o=son[o][wh];
        }
        size[o]++;
    }
    int ans;
    void query(int o,int dep,const string &s){
        if(o==0) return;
        if(dep==s.length()){
            ans+=size[o];
            return;
        }
        if(s[dep]=='#'||s[dep]=='@'){
            for(int j=0;j<26;j++) query(son[o][j],dep+1,s);
        }else{
            query(son[o][s[dep]-'a'],dep+1,s);
        }
    }
}trie;
int main(){
    ios::sync_with_stdio(false);
    int T; cin>>T;
    while(T--){
        int n,m;
        cin>>n>>m;
        rep(i,1,n) cin>>str[i];
        sort(str+1,str+1+n,cmp);
        string a,b;
        rep(i,1,m){
            cin>>a>>b;
            q[i]=Q(a,b,i);
        }
        sort(q+1,q+1+m);
        trie.init();
        int cur=n;
        string s;
        rrep(i,m,1){
            while(cur>0&&str[cur].length()>=q[i].len){
                trie.insert(str[cur]);
                cur--;
            }
            s.clear();
            int len=2*q[i].pre.length();
            int cur1=0,cur2=len>>1;
            for(int j=0;j<len;j++){
                char ch;
                if(j&1) ch=q[i].suf[--cur2];
                else ch=q[i].pre[cur1++];
                s.push_back(ch);
            }
            trie.ans=0; trie.query(1,0,s);
            ans[q[i].id]=trie.ans;
        }
        rep(i,1,m){
            cout<<ans[i]<<endl;
        }
    }
    return 0;
}