[Bzoj4943][Noi2017]蚯蚓（hash）

4943: [Noi2017]蚯蚓

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Time Limit: 50 Sec  Memory Limit: 512 MB
Submit: 237  Solved: 110
[Submit][Status][Discuss]

Description

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http://www.lydsy.com/JudgeOnline/upload/Noi2017D1.pdf

Input

Output

Sample Input

Sample Output

HINT

Source

分析：

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考试时秒切的一道题，看到子串长度<=50，考虑暴力hash。bzoj内存只有四分之一差评。

AC代码：（原内存可过）

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# include <iostream>
# include <cstdio>
# include <vector>
# include <cstring>
using namespace std;
typedef unsigned long long llu;
const int N = 5e5 + ;
const int M =  << ;
const int mod = ;
const llu bac = ;
int n,m,a[N],hs[M + ],mx,q[N][],nex[N],pre[N];
llu id[M + ],pw[],t[M + ];
char str[M + ];
vector<llu>Q[N];
int read()
{
    int x = ;char i = getchar();
    while(!isdigit(i))i = getchar();
    while(isdigit(i))x = x *  + i - '',i = getchar();
    return x;
}
int ins(llu s)
{
    int u = s % M;
    while(id[u])
    {
      if(id[u] == s)return u;
      u = (u + ) & (M - );
    }
    id[u] = s;
    return  u;
}
bool add(llu s,int x)
{
    int u = s % M;
    while(id[u])
    {
        if(id[u] == s)return hs[u] += x;
        u = (u + ) & (M - );
    }
    return ;
}
int lc[],rc[],lx,rx,p;llu g[];
void solve(int x,int y,int d)
{
    lx = rx = p = ;
    for(int i = ,u = x;i < mx && u;i++,u = pre[u])lc[++lx] = a[u];
    for(int i = ,u = y;i < mx && u;i++,u = nex[u])rc[++rx] = a[u];
    for(int i = lx;i >= ;i--)g[++p] = lc[i];
    for(int i = ;i <= rx;i++)g[++p] = rc[i];
    for(int i = ;i <= p;i++)g[i] = g[i - ] * bac + g[i];
    for(int i = ;i < lx;i++)
    {
        int r = min(i + mx,p);
        for(int j = lx + ;j <= r;j++)
        add(g[j] - g[i] * pw[j - i],d);
    }
}
int main()
{
    pw[] = ;for(int i = ;i <= ;i++)pw[i] = pw[i - ] * bac;
    n = read();m = read();
    for(int i = ;i <= n;i++)a[i] = read();
    for(int i = ;i <= m;i++)
    {
        q[i][] = read();
        if(q[i][] == )q[i][] = read(),q[i][] = read();
        else if(q[i][] == )q[i][] = read();
        else
        {
            scanf("%s",str + );int len = strlen(str + );
            int k  = read();
            mx = max(mx,k);
            for(int j = ;j <= len;j++)t[j] = t[j - ] * bac + str[j] - '';
            for(int j = k;j <= len;j++)
            Q[i].push_back(ins(t[j] - t[j - k] * pw[k]));
        }
    }
    for(int i = ;i <= n;i++)add(a[i],);
    for(int i = ;i <= m;i++)
    {
        int x,y;
        if(q[i][] == )
        {
            x = q[i][];y = q[i][];
            nex[x] = y;pre[y] = x;
            solve(x,y,);
        }
        else if(q[i][] == )
        {
           x = q[i][];y = nex[x];
           nex[x] = pre[y] = ;
           solve(x,y,-);
        }
        else
        {
            llu ans = ;
            for(int j = ;j < Q[i].size();j++)
            ans = ans * (llu)hs[Q[i][j]] % mod;
            printf("%d\n",(int)ans);
        }
    }
}