牛客多校2019-6C-Palindrome Mouse（回文自动机+统计支配点）

Palindrome Mouse

题意：若把给定字符串的所有不同回文子串放到一个集合内，求集合内有多少对回文子串满足其中一个是另一个的子串。

思路：

1. 建好回文自动机
2. 若设$an{s}_i$为回文树上能到达$i$节点的节点数（$0,1$除外），则题目要求的就是$\sum _{i=2}^{sz}an{s}_i$
3. 那么如何统计能到达某个节点的节点数呢？我们采用$DFS$的方法，一边沿着$ch$边跑，一边沿着$fail$边跑，并且将跑过的边打上标记，离开$DFS$的时候删除标记（用vector记录哪些点是刚刚打过标记的）
4. 然后，在每个节点处都累加答案即可

题目描述

Doctor has a string s consisting of only lowercase letters. Doctor has got the set of all palindromic substrings of a string s, denoted by the set S. Now, he wants to choose two distinct strings a and b from the set S which satisfy that the string a is a substring of the string b. How many different pairs (a, b) can Doctor choose?

输入描述:

There are multiple test cases. The first line contains an integer T (1≤T≤5), indicating the number of test cases. Test cases are given in the following.

Each test case consists of only one line, containing a non-empty string s (1≤∣s∣≤100000), consisting of only lowercase letters.

输出描述:

For each test case, output “Case #x: y” in one line (without quotes), where x indicates the case number starting from 1, and y denotes the answer to this test case.

示例

输入
4
aaaa
abba
abacaba
abc
输出
Case #1: 6
Case #2: 4
Case #3: 14
Case #4: 0

//#pragma comment(linker, "/STACK:102400000,102400000")
#include "bits/stdc++.h"
#define pb push_back
#define ls l,m,now<<1
#define rs m+1,r,now<<1|1
#define hhh printf("hhh\n")
#define see(x) (cerr<<(#x)<<'='<<(x)< pr;
inline int read() {int x=0;char c=getchar();while(c<'0'||c>'9')c=getchar();while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return x;}

const int maxn = 1e5+10;
const int mod = 1e9+7;
const double eps = 1e-9;

char s[maxn];
int ch[maxn][26], fail[maxn], len[maxn], cnt[maxn];
bool vis[maxn];
int last, sz, pos;
ll ans, res;

void init() {
	for(int i=0; i<=sz; ++i) {
		for(int j=0; j<26; ++j) ch[i][j]=0;
	}
	fail[0]=fail[1]=last=sz=1;
	len[1]=pos=-1;
	ans=0;
}

void add(int c) {
	++pos; int p=last;
	while(s[pos-len[p]-1]!=s[pos]) p=fail[p];
	if(!ch[p][c]) {
		int np=++sz, fp=fail[p];
		len[np]=len[p]+2;
		while(s[pos-len[fp]-1]!=s[pos]) fp=fail[fp];
		fail[np]=ch[fp][c];
		ch[p][c]=np;
	}
	last=ch[p][c];
}

void dfs(int u) {
	vector v;
	for(int i=u; i>1&&!vis[i]; i=fail[i]) vis[i]=1, res++, v.pb(i);
	ans+=u>1?res-1:0;
	for(int i=0; i<26; ++i) if(ch[u][i]) dfs(ch[u][i]);
	for(auto p: v) vis[p]=0, res--;
}

int main() {
	//ios::sync_with_stdio(false);
	int T=read();
	int t=0;
	while(T--) {
		scanf("%s", s);
		init();
		for(int i=0; s[i]; ++i) add(s[i]-'a');
		dfs(0), dfs(1);
		printf("Case #%d: %lld\n", ++t, ans);
	}
}