[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 * ], 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] = ;
     }

     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 ;
     }
 }T;
 #undef Next

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

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

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

 void get_hs(int mid, bool z)//获取所有长度<= 50的，跨mid的hs值
 {
     for(R i = ; i <= top; i ++) ls[i] = ls[i - ] * base + s[i];
     for(R i = ; i <= mid; i ++)//枚举开头
     {
         int lim1 = mid - i + , lim2 = top - i + ;//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 - ;
             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 = , tmp = ;//把j接到i之后
     Next[x] = y, last[y] = x, top = ;
     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, );
 }

 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 = , tmp = ;//把j接到i之后
     Next[x] = y, last[y] = x, top = ;
     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, );
 }

 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 = , tmp = ;//把i和它之后的一个蚯蚓断开
     last[y] = Next[x] = top = ;
     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, );
 }

 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 + ), top = strlen(ss + );
     int k = read(), b = top - k + ;
     for(R i = ; i <= top; i ++)
         ls[i] = ls[i - ] * base + ss[i] - '';
     LL ans = ;
     for(R i = ; i <= b; i ++)
     {
         us LL x = cal(i, i + k - );
         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 = ; i <= m; i ++)
     {
         int opt = read();
         if(opt == ) link();
         else if(opt == ) cut();
         else find();
     }
 }

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