HDU - 3948 The Number of Palindromes 后缀数组+ST表、distinct substring

The Number of Palindromes

 HDU
- 3948

Now, you are given a string S. We want to know how many distinct substring of S which is palindrome.

InputThe first line of the input contains a single integer T(T<=20), which indicates number of test cases. 

Each test case consists of a string S, whose length is less than 100000 and only contains lowercase letters. 

OutputFor every test case, you should output "Case #k:" first in a single line, where k indicates the case number and starts at 1. Then output the number of distinct substring of S which is palindrome. 

Sample Input

3
aaaa
abab
abcd

Sample Output

Case #1: 4
Case #2: 4
Case #3: 4

Source
HDU - 3948

My Solution题意：给出一个字符串，求出其回文子串的种数。后缀数组+ST表
把s变成 s + '#' + s'，其中s'表示s的反向串。所以当回文子串的长度为奇数时  [i+1,  _rank[n - sa[i] - 1]] 的height最小值表示以s[sa[i]]为中心的回文子串的最大半径(算上中心)； 长度为偶数时 [i+1,  _rank[n - sa[i]]] 的height最小值表示以s[sa[i]]为半径起点的回文子串的最大半径；
找这个最大半径，可以用ST表O(nlogn)的预处理出，然后每次O(1)的询问。所以跑出height以后，o，e分表表示长度为偶数和奇数的情况下，i-1时的最长半径长度，且o = min(o, height[i]), e = min(e, height[i])。此时 就是 ans += max(0, maxheight[] - o)； ans += max(0, maxheight)；然后刷新o,e。此外注意ST的查询操作要有 if(l > r) return 0;以保证每组构成回文子串的后缀只会查询一次。复杂度 O(nlogn)

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 2e5 + 8;
int sa[maxn], height[maxn];
int _rank[maxn], t1[maxn], t2[maxn], c[maxn];
string s;
inline void get_sa(const int &n, int m)
{
int i, k, *x = t1, *y = t2, p, j;
for(i = 0; i < m; i++) c[i] = 0;
for(i = 0; i < n; i++) ++ c[x[i] = s[i]];
for(i = 1; i < m; i++) c[i] += c[i - 1];
for(i = n - 1; i >= 0; i--) sa[-- c[x[i]]] = i;
for(k = 1; k <= n; k <<= 1){
p = 0;
for(i = n - k; i < n; i++) y[p ++] = i;
for(i = 0; i < n; i++) if(sa[i] >= k) y[p ++] = sa[i] - k;
for(i = 0; i < m; i++) c[i] = 0;
for(i = 0; i < n; i++) ++ c[x[y[i]]];
for(i = 1; i < m; i++) c[i] += c[i - 1];
for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
swap(x, y), p = 1, x[sa[0]] = 0;
for(i = 1; i < n; i++)
x[sa[i]] = (y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k]) ? p - 1 : p ++;
if(p >= n) break;
m = p;
}
k = 0;
for(i = 0; i < n; i++) _rank[sa[i]] = i;
for(i = 0; i < n; i++){
if(k) --k; if(!_rank[i]) continue;
j = sa[_rank[i] - 1];
while(s[i + k] == s[j + k]) k++;
height[_rank[i]] = k;
}
}

inline void print(const int &n)
{
for(int i = 1; i <= n; i++){
cout << i << " : " << _rank[sa[i]] << " " << sa[i] << endl;
for(int j = sa[i]; j < n; j++){
cout << s[j];
}
cout << endl;
}
cout << endl;
}

struct ST_list{
int stTable[maxn][32], preLog2[maxn];
inline void st_prepare(const int &n, int arr[]){
preLog2[1] = 0;
for(int i = 2; i <= n; i++){
preLog2[i] = preLog2[i-1];
if((1 << (preLog2[i] + 1)) == i){
preLog2[i]++;
}
}
for(int i = n - 1; i >= 0; i--){
stTable[i][0] = arr[i];
for(int j = 1; (i + (1 << j) - 1) < n; j++){
stTable[i][j] = min(stTable[i][j - 1], stTable[i + (1 << j - 1)][j - 1]); //1
}
}
}
inline int query_min(int l, int r){
if(l > r) return 0;
int len = r - l + 1, k = preLog2[len];
return min(stTable[l][k], stTable[r - (1 << k) + 1][k]);  //2
}
}st;

int main()
{
#ifdef LOCAL
freopen("9.in", "r", stdin);
//freopen("9.out", "w", stdout);
#endif // LOCAL
ios::sync_with_stdio(false); cin.tie(0);

int T, n, sz, i, o, e, ans, kase = 1;
cin >> T;
while(T--){
cin >> s;
sz = s.size();
s += '#';
for(i = sz - 1; i >= 0; i--) s += s[i];
n = sz + sz + 1;
get_sa(n+1, 256); st.st_prepare(n+1, height);
//print(n);
ans = 0; o = e = 0;
for(i = 2; i <= n; i++){
o = min(o, height[i]);
ans += max(0, st.query_min(i+1, _rank[n - sa[i] - 1]) - o); o = max(o, st.query_min(i+1, _rank[n - sa[i] - 1]));
//cout << i << " " << _rank[n - sa[i] - 1] << " " << st.query_min(i, _rank[n - sa[i] - 1]) << endl;
e = min(e, height[i]);
ans += max(0, st.query_min(i+1, _rank[n - sa[i]]) - e); e = max(e, st.query_min(i+1, _rank[n - sa[i]]));
//cout << o << " " << e << " " << ans << endl;
}
cout << "Case #" << kase << ": " << ans << "\n"; kase++;

}
return 0;
}

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