VK Cup 2012 Round2 E

Problem

给一个含有 N 个单词的词典，每个单词是 Si 。进行 M 次操作，操作分三种：一，删除词典里第 k 个单词，若已被删除则跳过；二，恢复词典里第 k 个单词，若已存在则跳过；三，给一篇文章 Article ，问当前词典里的单词在 Article 中出现的次数之和。

Limits

TimeLimit(ms):3000

MemoryLimit(MB):256

N,M∈[1,105]

k∈[1,N]

|Article|,|S1+S2+…+SN|∈[1,106]

字符集∈[a,z]

Look up Original Problem From here

Solution

AC自动机＋有向树＋树状数组优化

Complexity

TimeComplexity:O(|S|×log)

MemoryComplexity:O(|S|×26)

My Code

//Hello. I'm Peter.
#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cctype>
#include<ctime>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define peter cout<<"i am peter"<<endl
#define input freopen("data.txt","r",stdin)
#define randin srand((unsigned int)time(NULL))
#define INT (0x3f3f3f3f)*2
#define LL (0x3f3f3f3f3f3f3f3f)*2
#define gsize(a) (int)a.size()
#define len(a) (int)strlen(a)
#define slen(s) (int)s.length()
#define pb(a) push_back(a)
#define clr(a) memset(a,0,sizeof(a))
#define clr_minus1(a) memset(a,-1,sizeof(a))
#define clr_INT(a) memset(a,INT,sizeof(a))
#define clr_true(a) memset(a,true,sizeof(a))
#define clr_false(a) memset(a,false,sizeof(a))
#define clr_queue(q) while(!q.empty()) q.pop()
#define clr_stack(s) while(!s.empty()) s.pop()
#define rep(i, a, b) for (int i = a; i < b; i++)
#define dep(i, a, b) for (int i = a; i > b; i--)
#define repin(i, a, b) for (int i = a; i <= b; i++)
#define depin(i, a, b) for (int i = a; i >= b; i--)
#define pi 3.1415926535898
#define eps 1e-6
#define MOD 1000000007
#define MAXN 1001000
#define N 100100
#define M 26
ll ans;
int n,k,lens;
char s[MAXN];
bool here[N];

class Directed_Tree{
public:
    vector<int>vto[MAXN];
    inline void plant(int from,int to){
        vto[from].pb(to);
    }
    int num,mapto[MAXN],upto[MAXN],depth[MAXN];
    inline void dfs(int now,int kth){
        num+=1;
        mapto[now]=num;
        depth[num]=kth;
        int len=gsize(vto[now]);
        rep(i,0,len){
            dfs(vto[now][i],kth+1);
        }
        upto[now]=num;
    }
}dirtree;

class Bit{
public:
    ll c1[MAXN],c2[MAXN],sum[MAXN];
    int limit;
    inline int lowbit(int x){
        return x&-x;
    }
    inline void build_sum(int *a,int n){
        repin(i,1,n){
            sum[i]=sum[i-1]+a[i];
        }
    }
    inline void add(int x,ll val,ll *c){
        for(int i=x;i<=limit;i+=lowbit(i)){
            c[i]+=val;
        }
    }
    inline ll ask(int x,ll *c){
        ll res=0;
        for(int i=x;i>0;i-=lowbit(i)){
            res+=c[i];
        }
        return res;
    }
    inline void update(int l,int r,ll val){
        add(l,val,c1);
        add(r+1,-val,c1);
        add(l,l*val,c2);
        add(r+1,-(r+1)*val,c2);
    }
    inline ll query(int l,int r){
        ll res=sum[r]+(r+1)*ask(r,c1)-ask(r,c2);
        res-=sum[l-1]+l*ask(l-1,c1)-ask(l-1,c2);
        return res;
    }
}bit;

class Trie{
public:
    int num_node,trie[MAXN][M],val[MAXN],reval[N];
    inline int idx(char c){
        return c-'a';//attention!
    }
    inline void clear(){
        clr(trie[0]);
        num_node=1;
    }
    inline void insert(char *s,int len,int kth){
        int now=0,letter;
        rep(i,0,len){
            letter=idx(s[i]);
            if(!trie[now][letter]){
                trie[now][letter]=num_node;
                val[num_node]=0;
                clr(trie[num_node]);
                num_node+=1;
            }
            now=trie[now][letter];
        }
        val[now]=kth;
        reval[kth]=now;
    }
    int fail[MAXN],last[MAXN];
    queue<int>q;
    inline void bfs(){
        fail[0]=last[0]=0;
        int now=0;
        rep(i,0,M){
            int son=trie[now][i];
            if(son){
                fail[son]=0;
                last[son]=0;
                q.push(son);
            }
        }
        int son,f;
        while(!q.empty()){//bfs
            now=q.front();
            q.pop();
            if(val[now]) dirtree.plant(last[now],now);
            rep(i,0,M){
                son=trie[now][i];
                f=fail[now];
                if(!son){
                    trie[now][i]=trie[f][i];
                    continue;
                }
                while(f && !trie[f][i]) f=fail[f];
                fail[son]=trie[f][i];
                last[son]=val[fail[son]]?fail[son]:last[fail[son]];
                q.push(son);
            }
        }
    }
    inline void update_ans(int now){
        ans+=bit.query(dirtree.mapto[now],dirtree.mapto[now]);
    }
    inline void search(char *s,int len){
        int now=0,letter;
        ans=0;
        rep(i,0,len){
            letter=idx(s[i]);
            now=trie[now][letter];
            if(val[now]) update_ans(now);
            else if(last[now]) update_ans(last[now]);
        }
    }
}trie;

int main(){
    scanf("%d %d",&n,&k);
    trie.clear();
    repin(i,1,k){
        scanf("%s",s);
        lens=len(s);
        trie.insert(s,lens,i);
        here[i]=true;
    }
    trie.bfs();
    dirtree.dfs(0,0);
    bit.limit=dirtree.num;
    bit.build_sum(dirtree.depth,dirtree.num);
    repin(Case,1,n){
        char type;
        scanf(" %c",&type);
        if(type=='-'){
            int x;
            scanf("%d",&x);
            if(!here[x]) continue;
            here[x]=false;
            int node=trie.reval[x];
            bit.update(dirtree.mapto[node],dirtree.upto[node],-1);
        }
        else if(type=='+'){
            int x;
            scanf("%d",&x);
            if(here[x]) continue;
            here[x]=true;
            int node=trie.reval[x];
            bit.update(dirtree.mapto[node],dirtree.upto[node],+1);
        }
        else{//type=='?'
            scanf("%s",s);
            lens=len(s);
            trie.search(s,lens);
            printf("%lld\n",ans);
        }
    }
}