HDU 5157 Harry and magic string(回文树)

题意:

$N\le 10^5的字符串S,设T_1、T_2为S的2个回文子串,并且T_1和T_2不相交$
$求(T_1, T_2)的对数有多少$

分析:

$回文树预处理$
$pre[i]:=以i字符结尾的回文串的个数,suf[i]:=以i字符开头的回文串的个数$
$对中一个求前缀和,比如是suf[i],然后答案ans=\sum_{i=1}^{n-1}pre[i]\times sum[i+1]$

代码:

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//
// Created by TaoSama on 2016-04-09
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
struct PalindromicTree {
static const int M = 1e5 + 10, S = 26;
int n, sz, last;
int nxt[M][S], fail[M], len[M], s[M];
int cnt[M], dif[M];
int newNode(int l) {
len[sz] = l;
cnt[sz] = dif[sz] = 0;
memset(nxt[sz], 0, sizeof nxt[sz]);
return sz++;
}
void init() {
sz = last = 0;
newNode(0); newNode(-1);
s[n = 0] = -1; // 无关字符减少特判
fail[0] = 1;
}
int getFail(int u) {
while(s[n - len[u] - 1] != s[n]) u = fail[u];
return u;
}
int add(int c) {
s[++n] = c;
int u = getFail(last); // 找到这个回文串的匹配位置
int& v = nxt[u][c];
if(!v) {
int cur = newNode(len[u] + 2);
fail[cur] = nxt[getFail(fail[u])][c];
v = cur;
cnt[v] = cnt[fail[v]] + 1;
}
++dif[v];
last = v;
return cnt[v];
}
void count() {
//父亲累加儿子,如果 fail[v]=u ,则 u 一定是 v 的子回文串
for(int i = sz - 1; ~i; --i) dif[fail[i]] += dif[i];
}
} pt;
char s[N];
long long suf[N];
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%s", s + 1) == 1) {
int n = strlen(s + 1);
pt.init();
suf[n + 1] = 0;
for(int i = n; i; --i) {
int cnt = pt.add(s[i] - 'a');
suf[i] = suf[i + 1] + cnt;
}
pt.init();
long long ans = 0;
for(int i = 1; i <= n; ++i)
ans += pt.add(s[i] - 'a') * suf[i + 1];
printf("%I64d\n", ans);
}
return 0;
}


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