回文树模板

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;
}

求回文子串个数

如果是本质不同的，即就是回文树的点数减去两个根
否则就是 (fail) 树上的 (deep) 和也就是 (size) 和这两个显然相等

一个点的 (deep) 表示以这个点包含的端点集合为右端点的回文串的个数
一个点的 (size) 表示这个点表示的回文串出现的端点集合大小

SPOJ NUMOFPAL

# 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(1005);

int fa[maxn], num[maxn], len[maxn], son[26][maxn], ans, n, tot, last;
char s[maxn];

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 q = fa[p], np = ++tot;
		while(s[pos - len[q] - 1] != s[pos]) q = fa[q];
		fa[np] = son[c][q];
		son[c][p] = np, len[np] = len[p] + 2;
	}
	last = son[c][p], ++num[last];
}

int main(){
	scanf(" %s", s + 1), n = strlen(s + 1);
	tot = 1, len[1] = -1, fa[0] = fa[1] = 1;
	for(RG int i = 1; i <= n; ++i) Extend(i, s[i] - 'a');
	for(RG int i = tot; i; --i) num[fa[i]] += num[i];
	for(RG int i = 2; i <= tot; ++i) ans += num[i];
	printf("%d
", ans);
	return 0;
}

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