2019暑期杭电多校HDU6599 I Love Palindrome String

2019杭电多校HDU6599 I Love Palindrome String

题目链接

I Love Palindrome String
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 582 Accepted Submission(s): 220
Problem Description
You are given a string S=s1s2…s|S| containing only lowercase English letters. For each integer i∈[1,|S|] , please output how many substrings slsl+1…sr satisfy the following conditions:
∙ r−l+1 equals to i.
∙ The substring slsl+1…sr is a palindrome string.
∙ slsl+1…s⌊(l+r)/2⌋ is a palindrome string too.
|S| denotes the length of string S.
A palindrome string is a sequence of characters which reads the same backward as forward, such as madam or racecar or abba.

Input
There are multiple test cases.
Each case starts with a line containing a string S(1≤|S|≤3×105) containing only lowercase English letters.
It is guaranteed that the sum of |S| in all test cases is no larger than 4×106.

Output
For each test case, output one line containing |S| integers. Any two adjacent integers are separated by a space.

Sample Input
abababa

Sample Output
7 0 0 0 3 0 0

Source
2019 Multi-University Training Contest 2

题意：
给你一个字符串，求对于每一个长度i-len，问有多少个，回文串的前一半也是回文串
思路：
回文树裸题，用hash判断前后两部分是否相同

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <string>
#include <iostream>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define pd(n) printf("%d\n", (n))
#define pdd(n,m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n,m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define mm(a,n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define mod(x) ((x)%MOD)
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const int MAXN = 100005 ;
const int N = 3e5+10;
const ull hash1 = 201326611;
const ull hash2 = 50331653;
inline int read()
{int ret = 0, sgn = 1;char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-') sgn = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){ret = ret*10 + ch - '0';ch = getchar();}return ret*sgn;
}
char str[N];
ll ret[N];
ull ha[N], pp[N];
ull getha(int l, int r)
{if (l == 0) return ha[r];return ha[r] - ha[l - 1] * pp[r - l + 1];
}
bool check(int l, int r)
{int len = r - l + 1;int mid = (l + r) >> 1;if (len & 1) return getha(l, mid) == getha(mid, r);else return getha(l, mid) == getha(mid + 1, r);
}
struct PTree
{int nxt[N][26]; //i节点对应的回文串在两边各加一个字符后变成的节点编号int fail[N]; //fail[i]表示的串是i的最长后缀回文串int cnt[N]; //表示这个串出现过几次int num[N]; //节点i这个串的后缀有多少是回文的int len[N]; //节点i表示的回文串的长度int S[N]; //S[i]是第i次添加的字符int last; //以n结束的最长回文串所在的节点int n; //目前添加的字符个数int p; //下一个新建节点的标号int id[N];inline int newnode(int l){//新建节点for(int i = 0; i < 26; ++i) nxt[p][i] = 0;cnt[p] = num[p] = 0;len[p] = l;return p++;}inline void init(){p = 0;newnode(0);newnode(-1);last = n = 0;S[n] = -1;fail[0] = 1;}inline int get_fail(int x){while(S[n - len[x] - 1] != S[n]) x = fail[x];return x;/*新加进来一个字符时，只有Str[Nnow−len[x]−1]=Str[Nnow]Str[Nnow−len[x]−1]=Str[Nnow]我们才可以说是能构成一个新的回文串（构成回文节点唯一方法，就是在已有基础上找一个节点，在其两端各拓展一个字符，也就是我们所说的找到一个x，使式子成立）。于是我们就可以去不停给last跑fail，直到满足条件。*/}inline void add(int c)//加一个字符{c-='a';S[++n] = c;int cur = get_fail(last); //通过上一个回文串找这个串的匹配位置if(!nxt[cur][c]){//这个回文串从未出现过int now = newnode(len[cur] + 2);fail[now] = nxt[get_fail(fail[cur])][c]; //和AC自动机简直不要太类似nxt[cur][c] = now;num[now] = num[fail[now]] + 1;}last = nxt[cur][c];++cnt[last];id[last] = n;//}inline void count1()//计数{for(int i = p - 1; i >= 0; --i) cnt[fail[i]]+= cnt[i];for (int i = 2; i < p; i++){///cout << id[i] - len[i] << " " << id[i] - 1 << endl;if (check(id[i] - len[i], id[i] - 1)){ret[len[i]] += cnt[i];}}}inline void work(){pp[0] = 1;for (int i = 1; i < N; i++){pp[i] = hash1 * pp[i - 1];}memset(ret, 0, sizeof(ret));init();int slen = strlen(str);ha[0] = str[0];for (int i = 0; i < slen; i++){add(str[i]);}for (int i = 1; i < slen; i++){ha[i] = ha[i - 1] * hash1 + str[i];}count1();printf("%lld", ret[1]);for (int i = 2; i <= slen; i++){printf(" %lld", ret[i]);}printf("\n");}} pam ;
int main()
{while(~ss(str))pam.work();return 0;
}

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