Bzoj2555: SubString

题面

传送门

Sol

考虑求每个串在模板串中出现的次数
就在\(sam\)上走就行了,因为它的每一条路径都是它的一个子串
走到最后一个点,若匹配,那么它的答案就是\(parent\)树的这个点的子树大小

然后带修改就写个\(LCT\)维护\(parent\)树就好了
\(LCT\)维护子树信息,非常好写

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
 
IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}
 
const int maxn(1200005);
 
struct LCT{
    int fa[maxn], val[maxn], sum[maxn], ch[2][maxn];
     
    IL int Son(RG int x){
        return ch[1][fa[x]] == x;
    }
     
    IL int Isroot(RG int x){
        return ch[0][fa[x]] != x && ch[1][fa[x]] != x;
    }
     
    IL void Update(RG int x){
        sum[x] = val[x] + sum[ch[0][x]] + sum[ch[1][x]];
    }
     
    IL void Rotate(RG int x){
        RG int y = fa[x], z = fa[y], c = Son(x);
        if(!Isroot(y)) ch[Son(y)][z] = x; fa[x] = z;
        ch[c][y] = ch[!c][x], fa[ch[c][y]] = y;
        ch[!c][x] = y, fa[y] = x;
        Update(y);
    }
     
    IL void Splay(RG int x){
        for(RG int y = fa[x]; !Isroot(x); Rotate(x), y = fa[x])
            if(!Isroot(y)) Son(x) ^ Son(y) ? Rotate(x) : Rotate(y);
        Update(x);
    }
     
    IL void Access(RG int x){
        for(RG int y = 0; x; y = x, x = fa[x]){
            Splay(x);
            val[x] += sum[ch[1][x]] - sum[y];
            ch[1][x] = y, Update(x);
        }
    }
     
    IL void Cut(RG int x){
        Access(x), Splay(x);
        ch[0][x] = fa[ch[0][x]] = 0, Update(x);
    }
     
    IL void Link(RG int x, RG int y){
        Splay(x), Access(y), Splay(y);
        fa[x] = y, val[y] += sum[x], Update(y);
    }
} lct;
 
int tot = 1, last = 1, trans[26][maxn], fa[maxn], len[maxn];
 
IL void Extend(RG int c){
    RG int p = last, np = ++tot;
    last = np, len[np] = len[p] + 1, lct.val[np] = 1;
    while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
    if(!p) fa[np] = 1, lct.Link(np, 1);
    else{
        RG int q = trans[c][p];
        if(len[q] == len[p] + 1) fa[np] = q, lct.Link(np, q);
        else{
            RG int nq =  ++tot;
            lct.Link(nq, fa[q]);
            fa[nq] = fa[q], len[nq] = len[p] + 1;
            for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
            if(fa[q]) lct.Cut(q);
            lct.Link(q, nq), lct.Link(np, nq);
            fa[q] = fa[np] = nq;
            while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
        }
    }
}
 
int q, mask, ans;
char s[maxn * 3], op[8];
 
IL void NewOption(RG int l){
        RG int seed = mask;
    for(RG int i = 0; i < l; ++i){
            seed = (seed * 131 + i) % l;
        swap(s[i], s[seed]);
    }
}
 
int main(RG int argc, RG char *argv[]){
    q = Input(), scanf(" %s", s);
    for(RG int i = 0, l = strlen(s); i < l; ++i) Extend(s[i] - 'A');
    for(RG int i = 1; i <= q; ++i){
        scanf(" %s %s", op, s);
        RG int l = strlen(s);
        NewOption(l);
        if(op[0] == 'Q'){
            RG int nw = 1;
            for(RG int i = 0; i < l; ++i) nw = trans[s[i] - 'A'][nw];
            if(!nw) ans = 0;
            else lct.Access(nw), lct.Splay(nw), ans = lct.val[nw];
            printf("%d\n", ans), mask ^= ans;
        }
        else for(RG int i = 0; i < l; ++i) Extend(s[i] - 'A');
    }
    return 0;
}

posted @ 2018-04-23 19:20  Cyhlnj  阅读(102)  评论(0编辑  收藏  举报