回文树模板

Posted 2020-11-13 cjoieryl

存代码
学习的博客
然后国家集训队2017年的论文

在后面插入的

IL void Extend(RG int pos, RG int c){
    RG int p = last;
    while(s[pos - len[p] - 1] != s[pos]) p = fa[p];
    if(!son[c][p]){
        RG int np = ++tot, q = fa[p];
        while(s[pos - len[q] - 1] != s[pos]) q = fa[q];
        len[np] = len[p] + 2, fa[np] = son[c][q];
        son[c][p] = np;
    }
    last = son[c][p], ++size[last];
}

支持前后插入，维护最长回文前缀和最长回文后缀
前缀的(fail)和后缀的(fail)相同，因为回文串的对称性
题目vjudge/HDU：Victor and String

# include <bits/stdc++.h>
# define IL inline
# define RG register
# 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(1e5 + 5);

int n, tot, prel, sufl, son[26][maxn], len[maxn], deep[maxn], fa[maxn], l, r;
ll ans;
char s[maxn << 1];

IL void Init(){
    for(RG int i = l; i <= r; ++i) s[i] = 'z' + 1;
    for(RG int i = 0; i <= tot; ++i){
        len[i] = deep[i] = fa[i] = 0;
        for(RG int j = 0; j < 26; ++j) son[j][i] = 0;
    }
    prel = sufl = 0, tot = 1, fa[1] = fa[0] = 1, len[1] = -1;
}

IL void Extend(RG int pos, RG int c, RG int &last, RG int op){
    RG int p = last;
    while(s[pos - op * (len[p] + 1)] != s[pos]) p = fa[p];
    if(!son[c][p]){
        RG int np = ++tot, q = fa[p];
        while(s[pos - op * (len[q] + 1)] != s[pos]) q = fa[q];
        len[np] = len[p] + 2, fa[np] = son[c][q];
        son[c][p] = np;
    }
    last = son[c][p], deep[last] = deep[fa[last]] + 1, ans += deep[last];
    if(len[last] == r - l + 1) prel = sufl = last;
}

int main(){
    while(scanf("%d", &n) != EOF){
        Init(), l = 1e5, r = l - 1, ans = 0;
        for(RG int i = 1; i <= n; ++i){
            RG int op = Input();
            if(op <= 2){
                RG char c; scanf(" %c", &c);
                if(op == 1) s[--l] = c, Extend(l, c - 'a', prel, -1);
                else s[++r] = c, Extend(r, c - 'a', sufl, 1);
            }
            else printf("%lld
", op == 3 ? tot - 1 : ans);
        }
    }
    return 0;
}

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