SPOJ Number of Palindromes(回文树)

来源：互联网发布：最好的建筑设计软件编辑：程序博客网时间：2024/04/30 00:55

Number of Palindromes

Time Limit: 100MS Memory Limit: 1572864KB 64bit IO Format: %lld & %llu

Submit Status

Description

Each palindrome can be always created from the other palindromes, if a single character is also a palindrome. For example, the string "malayalam" can be created by some ways:

* malayalam = m + ala + y + ala + m
* malayalam = m + a + l + aya + l + a + m

We want to take the value of function NumPal(s) which is the number of different palindromes that can be created using the string S by the above method. If the same palindrome occurs more than once then all of them should be counted separately.

Input

The string S.

Output

The value of function NumPal(s).

Limitations

0 < |s| <= 1000

Example

Input:

malayalam

Output:

Hint

Added by:The quick brown fox jumps over the lazy dogDate:2010-10-18Time limit:0.100s-0.170sSource limit:50000BMemory limit:1536MBCluster:Cube (Intel G860)Languages:AllResource:Udit Agarwal

回文树：

参考大牛的博客：http://blog.csdn.net/lwfcgz/article/details/48739051

回文树是一种处理回文串的强大工具

#include <iostream>#include <string.h>#include <stdlib.h>#include <algorithm>#include <math.h>#include <stdio.h>using namespace std;#define MAX 1005struct Node{    int next[26];    int len;    int sufflink;    int num;}tree[MAX];char s[MAX];int num;int suff;bool addLetter(int pos){    int cur=suff,curlen=0;    int let=s[pos]-'a';    while(1)    {        curlen=tree[cur].len;        if(pos-1-curlen>=0&&s[pos-1-curlen]==s[pos])            break;        cur=tree[cur].sufflink;    }    if(tree[cur].next[let])    {        suff=tree[cur].next[let];        return false;    }    num++;    suff=num;    tree[num].len=tree[cur].len+2;    tree[cur].next[let]=num;    if(tree[num].len==1)    {        tree[num].sufflink=2;        tree[num].num=1;        return true;    }    while(1)    {        cur=tree[cur].sufflink;        curlen=tree[cur].len;        if(pos-1-curlen>=0&&s[pos-1-curlen]==s[pos])        {            tree[num].sufflink=tree[cur].next[let];            break;        }    }    tree[num].num=1+tree[tree[num].sufflink].num;    return true;}void initTree(){    num=2;suff=2;    tree[1].len=-1;tree[1].sufflink=1;    tree[2].len=0;tree[2].sufflink=1;}int main(){    scanf("%s",s);    int len=strlen(s);    initTree();    long long int ans=0;    for(int i=0;i<len;i++)    {         addLetter(i);         ans+=tree[suff].num;    }    printf("%d\n",ans);    return 0;}