最长回文
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18219 Accepted Submission(s): 6689
Problem Description
给出一个只由小写英文字符a,b,c...y,z组成的字符串S,求S中最长回文串的长度. 回文就是正反读都是一样的字符串,如aba, abba等
Input
输入有多组case,不超过120组,每组输入为一行小写英文字符a,b,c...y,z组成的字符串S 两组case之间由空行隔开(该空行不用处理) 字符串长度len <= 110000
Output
每一行一个整数x,对应一组case,表示该组case的字符串中所包含的最长回文长度.
Sample Input
aaaa
abab
Sample Output
4
3
Source
1 //2016.10.06 2 #include3 #include 4 #include 5 6 using namespace std; 7 8 const int N = 110005; 9 char str[N], s[N<<1];10 int a[N<<1];11 12 13 //s为待求字符串,len为字符串长度,a[i]为以s[i]为中心的最长回文串的长度的一半。若s为前后添加过特殊字符的字符串,则返回值为原来字符串种最长回文串的长度。14 int manacher(char *s, int *a, int len)15 {16 a[0] = 0;17 int ans = 0, j;18 for(int i = 0; i < len; )19 {20 while(i-a[i]>0 && s[i+a[i]+1]==s[i-a[i]-1])21 a[i]++;22 if(ans < a[i])ans = a[i];23 j = i+1;24 while(j<=i+a[i] && i-a[i]!=i+i-j-a[i+i-j]){25 a[j] = min(a[i+i-j], i+a[i]-j);26 j++;27 }28 a[j] = max(i+a[i]-j, 0);29 i = j;30 }31 return ans;32 }33 34 int main()35 {36 int len;37 while(scanf("%s", str)!=EOF)38 {39 len = 2*strlen(str)+1;40 for(int i = 0; str[i] != '\0'; i++)//将字符串每一位中间插入一个特殊字符41 {42 s[i+i] = '\0';43 s[i+i+1] = str[i];44 }45 s[len-1] = '\0';46 printf("%d\n", manacher(s, a, len));47 }48 49 return 0;50 }