最长回文 Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 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 2009 Multi-University Training Contest 16 - Host by NIT Recommend lcy | We have carefully selected several similar problems for you: 1358 1686 3336 3065 3746
#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; #define maxn 20000050 char s[maxn],str[2*maxn]; int p[maxn]; void Manacher(int *p,char *str,int len){ int mx=0,idx=0; for(int i=1;i<=len;i++){ p[i] = mx>i ? min(p[2*idx-i],mx-i) : 1; while(str[i+p[i]]==str[i-p[i]]) p[i]++; if(i+p[i] > mx){ mx=i+p[i]; idx=i; } } } int main(){ while(scanf("%s",s)!=EOF){ int nn=strlen(s); int n=nn*2+2; str[0]='$'; for(int i=0;i<=nn;i++){ str[i*2+1]='#'; str[i*2+2]=s[i]; } Manacher(p,str,n); int ans=1; for(int i=0;i<n;i++) ans=max(ans,p[i]); printf("%d\n",ans-1); } return 0; }Manacher算法裸题
