HDU 3608 最長迴文(擴充KMP)

最長迴文

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5319    Accepted Submission(s): 1815


Problem Description給出一個只由小寫英文字元a,b,c...y,z組成的字串S,求S中最長迴文串的長度.
迴文就是正反讀都是一樣的字串,如aba, abba等 

Input輸入有多組case,不超過120組,每組輸入為一行小寫英文字元a,b,c...y,z組成的字串S
兩組case之間由空行隔開(該空行不用處理)
字串長度len <= 110000 

Output每一行一個整數x,對應一組case,表示該組case的字串中所包含的最長迴文長度. 

Sample Input

aaaaabab

 

Sample Output

43

 

Source2009 Multi-University Training Contest 16 - Host
by NIT 

Recommendlcy 這個題目我是參考這篇部落格寫的！http://www.cnblogs.com/chenlie/archive/2011/05/30/2063100.html這個擴充的KMP本來就不是怎麼好理解，花了半天時間把擴充的KMP搞懂了，想應用一下，沒想到想用他求一個最長迴文字串都這麼困難，表示這個演算法不好理解啊擴充KMP演算法可以看我轉的這篇部落格http://chenhongyu940407.blog.163.com/blog/static/20500925420136251431685/講的非常詳細，思路非常清晰擴充KMP的最佳化思路和KMP基本上是一樣的就是計算過的不再計算！

#include <iostream>#include <cstdio>#include <string>#include <cstring>using namespace std;#define MAX 110010int next1[MAX];int next2[MAX];int next[MAX];int ans;char a[MAX];char b[MAX];int getNext1(int n){    int i=0,j=0,k=0,p=0,a=0,l=0;    next[0]=n;    while(j+1 < n && b[j+1]==b[j]) j++;    next[1]=j;    a=1;    for(i=2;i<n;i++)    {        p=a+next[a]-1;        l=next[i-a];        if(i+l <= p)        next[i]=l;        else        {            j=p-i+1;            if(j<0)            j=0;            while(i+j < n && b[j]==b[i+j]) j++;            next[i]=j;            a=i;        }    }    return 0;}void getNext2(char *s, int *nexts, int n, int m){     getNext1(m);     int i, j, k;     j = 0;     k = 0;     while (s[j] == b[j])j++;     nexts[0] = j;     for (i = 1; i < n; i++)     {         if (next[i-k] + i < nexts[k] + k)             nexts[i] = next[i-k];         else         {             j = nexts[k] + k - i;             if (j < 0) j = 0;             while (i + j < n && s[i+j] == b[j])j++;             nexts[i] = j;             k = i;         }     }}void res(char *s, int n){     char st;     int len = (n>>1);     for (int i = 0; i < len; i++)     {         st = s[i];         s[i] = s[n-i-1];         s[n-i-1] = st;     }}void find(char *s, int n){     if (ans >= n || n < 2)return;     int i, k, x;     int mid;     mid = (n>>1);     for (i = mid; i < n; i++)b[i-mid] = s[i];     b[i-mid] = 0;     res(s, n);     getNext2(s, next1, n, i-mid);     res(s, n);     for (i = 0; i < mid; i++)       b[i] = s[mid-i-1];     b[i] = 0;     getNext2(s, next2, n, mid);     next1[n] = next2[n] = 0;     for (i = 0; i < mid; i++)     {         if (next2[i] >= mid - i)         {             x = mid - i + 2 * next1[n-i];             if (ans < x)ans = x;         }     }     for (i = mid; i < n; i++)     {         if (next1[n-i] >= i - mid)         {             x = i - mid + 2*next2[i];             if (ans < x) ans = x;         }     }     find(s, mid);     find(s+mid, n-mid);}int main(){    int  n;    while (scanf("%s", a) != EOF)    {          n = strlen(a);          ans = 1;          find(a, n);          printf("%d\n", ans);    }    return 0;}