[NOI2017]蚯蚓排队 hash

题面：洛谷

题解：

　　我们暴力维护当前所有队伍内的所有子串（长度k = 1 ~ 50)的出现次数。

　　把每个子串都用一个hash值来表示，每次改变队伍形态都用双向链表维护，并暴力更新出现次数。

　　现在考虑复杂度。

　　如果只有连接，没有断开，那么复杂度为不同的子串个数：50n(注意只要O(n)预处理前缀和后缀hash就可以做到O(1)得到一个子串的hash值)

　　如果有断开，那么最多就每断开一次就对应要再重连一次。所以复杂度最多增加2500 * 断开次数，而断开次数1e3....

　　所以复杂度就是正确的了。

　　此题略卡常。

　　如果T了一两个点，，，就多交几次吧，，，说不定哪次就过了呢？

#include<bits/stdc++.h>
using namespace std;
#define R register int
#define AC 201000
#define ac 10501000
#define LL long long
#define us unsigned
#define maxn 49//每边最多49个，以为另外一边至少一个
#define base 19260817//2333//直接用10？？？
#define bu 16777215//49999991//100000081//cket 一个质数，，，
#define mod 998244353
#define h(f) (f & bu)

int n, m, top;
int v[AC], last[AC], Next[AC];
us LL hs[AC], ls[ac], p[ac];
int Head[ac * 3], Next1[ac], date[ac], tot, id;
int cnt[ac]; us LL power[ac];//存下每个id对应的hash值以及出现次数
int s[AC];
char ss[ac];
//int tot, id;

#define Next Next1
//inline int h(int f){
    //return (((f & bu) ^ (mod >> 5)) + 1);
//}

struct node{
    
    inline void add(int f, us LL x)
    {//如果x这个表上没有k这个值，那就要新开id，否则直接用原来的id
    //    printf("%d %llu\n", f, x);
    //    printf("%llu\n", x);
        for(R i = Head[f]; i; i = Next[i])
            if(power[date[i]] == x) {++ cnt[date[i]]; return ;}
        int tmp = ++ id;
        date[++ tot] = tmp, Next[tot] = Head[f], Head[f] = tot;
        power[tmp] = x, cnt[tmp] = 1;
    }
    
    inline void del(int f, us LL x)//找到这个值并删除一个
    {
        for(R i = Head[f]; i; i = Next[i])
            if(power[date[i]] == x) {-- cnt[date[i]]; return ;}
    }
    
    inline int find(int f, us LL x)
    {
        for(R i = Head[f]; i; i = Next[i])
            if(power[date[i]] == x) return cnt[date[i]];
        return 0;
    }
}T;
#undef Next

inline int read()
{
    int x = 0;char c = getchar();
    while(c > '9' || c < '0') c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x;
}

void pre()
{
    n = read(), m = read(), p[0] = 1;
    for(R i = 1; i <= n; i ++) v[i] = read(), T.add(h(v[i]), v[i]);//先把单个的加进去
    for(R i = 1; i <= n; i ++) p[i] = p[i - 1] * base;//自然溢出
}

us LL cal(int l, int r){//返回[l, r]的hash值
    return ls[r] - ls[l - 1] * p[r - l + 1];
}

void get_hs(int mid, bool z)//获取所有长度<= 50的，跨mid的hs值
{
    for(R i = 1; i <= top; i ++) ls[i] = ls[i - 1] * base + s[i];
    for(R i = 1; i <= mid; i ++)//枚举开头
    {
        int lim1 = mid - i + 2, lim2 = top - i + 1;//长度要在[lim1, lim2]的范围内
        for(R len = lim1; len <= lim2; len ++)//才能保证合法
        {
            int r = i + len - 1;
            us LL x = cal(i, r);//获取区间hash值
            if(z) T.add(h(x), x);
            else T.del(h(x), x);
        }
    }
}

void link1()//每次合并的时候暴力加hash值，每次最多增加1250个
{
    int x = read(), y = read(), cnt = 0, tmp = 0;//把j接到i之后
    Next[x] = y, last[y] = x, top = 0;
    for(R i = x; i && cnt < maxn; i = last[i]) ++ cnt, tmp = i;
    for(R i = tmp; i != x; i = Next[i]) s[++ top] = v[i];
    s[++ top] = v[x], cnt = top;
    for(R i = y; i && top - cnt != maxn; i = Next[i]) s[++ top] = v[i];
    get_hs(cnt, 1);
}

void link()//每次合并的时候暴力加hash值，每次最多增加1250个
{
    int x = read(), y = read(), cnt = 0, tmp = 0;//把j接到i之后
    Next[x] = y, last[y] = x, top = 0;
    for(R i = x; i && cnt < maxn; i = last[i]) ++ cnt, tmp = i;
    for(R i = tmp; i != x; i = Next[i]) s[++ top] = v[i];
    s[++ top] = v[x], cnt = top;
    for(R i = y; i && top - cnt != maxn; i = Next[i]) s[++ top] = v[i];
    get_hs(cnt, 1);
}

void cut()//每次合并的时候暴力减hash值，每次最多减少1250个,但总体很小
{
    int x = read(), y = Next[x], cnt = 0, tmp = 0;//把i和它之后的一个蚯蚓断开
    last[y] = Next[x] = top = 0;
    for(R i = x; i && cnt < maxn; i = last[i]) ++ cnt, tmp = i;
    for(R i = tmp; i != x; i = Next[i]) s[++ top] = v[i];
    s[++ top] = v[x], cnt = top;
    for(R i = y; i && top - cnt != maxn; i = Next[i]) s[++ top] = v[i];
    get_hs(cnt, 0);
}

void find()//处理hash值的时候暴力O(n)处理前缀hash值，然后查询的时候也O(n)遍历查询
{//因为s长度之和最多1e7....
    //cin >> ss + 1, top = strlen(ss + 1);
    scanf("%s", ss + 1), top = strlen(ss + 1);
    int k = read(), b = top - k + 1;
    for(R i = 1; i <= top; i ++) 
        ls[i] = ls[i - 1] * base + ss[i] - '0';    
    LL ans = 1;
    for(R i = 1; i <= b; i ++)
    {
        us LL x = cal(i, i + k - 1);
        ans *= T.find(h(x), x);
        //printf("!!%d ", T.find(x & bu, x));
        if(ans > mod) ans %= mod;
    }
    //printf("\n");
    printf("%lld\n", ans);
}

void work()
{
    for(R i = 1; i <= m; i ++)
    {
        int opt = read();
        if(opt == 1) link();
        else if(opt == 2) cut();
        else find();
    }
}

int main()
{
//    freopen("in.in", "r", stdin);
    pre();
    work();
    //fclose(stdin);
    return 0;
}