hdu 5157 回文树

Harry and magic string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 147    Accepted Submission(s): 72

Problem Description

Harry got a string T, he wanted to know the number of T’s disjoint palindrome substring pairs. A string is considered to be palindrome if and only if it reads the same backward or forward. For two substrings of T:x=T[a1…b1],y=T[a2…b2](where a1 is the beginning index of x,b1 is the ending index of x. a2,b2 as the same of y), if both x and y are palindromes and b1<a2 or b2<a1 then we consider (x, y) to be a disjoint palindrome substring pair of T.

 

Input

There are several cases.
For each test case, there is a string T in the first line, which is composed by lowercase characters. The length of T is in the range of [1,100000].

 

Output

For each test case, output one number in a line, indecates the answer.

 

Sample Input

acaaaaa

 

Sample Output

315
Hint
For the first test case there are 4 palindrome substrings of T.They are:S1=T[0,0]S2=T[0,2]S3=T[1,1]S4=T[2,2]And there are 3 disjoint palindrome substring pairs.They are:(S1,S3) (S1,S4) (S3,S4).So the answer is 3.
 

 

Source

BestCoder Round #25

 

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反着构造一遍回文树，通过每个节点的后缀包含的回文串数目（称其为prenum值）求前缀和得到反向字符串的前i位包含的回文串数目，再正着构造一遍回文树，在用每个节点的prenum值乘上这个节点右侧的回文串数目，求和即可。

#include <iostream>#include <cstdio>#include <map>#include <cstring>#include <string>#include <queue>#define M 110#define N 110000using namespace std;class Node{public:    int pre, next[30], len, prenum, num;    void setValue(int _len)    {        len = _len;        pre = 0;        memset(next, 0, sizeof next);        prenum = 0;        num = 0;    }};Node node[N];char str[N], s[N];int preNum[2][N];long long preSum[N];class PaliTree{public:    int cnt, root0, root1, last, n;    int newNode(int _len)    {        node[++cnt].setValue(_len);        return cnt;    }    void init()    {        cnt = -1; n = 0;        str[0] = -1;        root0 = newNode(0);        root1 = newNode(-1);        node[root0].pre = root1;        node[root1].pre = root0;        last = root0;    }    int nextNode(int p)    {        for(; str[n-node[p].len-1] != str[n]; p = node[p].pre);        return p;    }    void extend(char cc, int* a)    {        int ch = cc-'a';        str[++n] = ch;        int p = nextNode(last);        int nxt = node[p].next[ch];        if(!nxt)        {            nxt = newNode(node[p].len+2);            node[nxt].pre = node[nextNode(node[p].pre)].next[ch];            node[p].next[ch] = nxt;            node[nxt].prenum = node[node[nxt].pre].prenum+1;        }        last = nxt;        node[last].num++;        a[n] = node[last].prenum;    }};PaliTree tree;int main(){//freopen("C:\\Users\\zfh\\Desktop\\in.txt", "r", stdin);    while(scanf(" %s", s) != -1)    {        tree.init();        int len = strlen(s);        for(int i = 0; i < len; i++)        {            tree.extend(s[i], preNum[0]);        }        tree.init();        for(int i = len-1; i >= 0; i--)        {            tree.extend(s[i], preNum[1]);        }        preSum[1] = preNum[1][1];        for(int i = 2; i <= len; i++)        {            preSum[i] = preSum[i-1]+preNum[1][i];        }        long long ans = 0;        for(int i = 1; i <= len; i++)        {            ans += preNum[0][i]*preSum[len-i];        }        printf("%I64d\n", ans);    }    return 0;}