hdu3068
最长回文
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17251 Accepted Submission(s): 6351
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
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
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
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题解:
manacher算法的裸体
AC代码:
#include<cstdio> #include<cstring> #include<iostream> using namespace std; #define N 2000100 int p[N]; char s[N],S[N]; int manacher(int len){ int ans=0,id=0,mx=0; for(int i=1;i<len;i++){ p[i]=mx>i?min(p[id+id-i],mx-i):1; while(S[i-p[i]]==S[i+p[i]]) p[i]++; if(i+p[i]>mx) mx=i+p[i],id=i; ans=max(ans,p[i]-1); } return ans; } void deal(int len){ int l=0; S[l++]='$';S[l++]='#'; for(int i=0;i<len;i++) S[l++]=s[i],S[l++]='#'; S[l++]='\0'; } int main(){ //freopen("sh.txt","r",stdin); while(scanf("%s",s)==1){ memset(S,0,sizeof S); memset(p,0,sizeof p); int len=strlen(s); deal(len); printf("%d\n",manacher(len+len+2)); //fill(S,S+len+len+2,0); //fill(p,p+len+len+2,0); } return 0; }
A2:
#include<cstdio> #include<cstring> #include<iostream> using namespace std; #define N 210100 int p[N],l; char s[N],S[N]; int manacher(){ int ans=0,mx=-1,id=-1; for(int i=1;i<=l;i++){ if(id+mx>=i) p[i]=min(p[(id<<1)-i],id+mx-i); while(i-p[i]-1>=0&&i+p[i]+1<=l&&S[i-p[i]-1]==S[i+p[i]+1]) p[i]++; if(i+p[i]>id+mx) mx=p[i],id=i; ans=max(ans,p[i]); } return ans; } int main(){ while(scanf("%s",s)==1){ memset(S,0,sizeof S); memset(p,0,sizeof p); int len=strlen(s); l=-1; for(int i=0;i<len;i++) S[++l]='#', S[++l]=s[i]; S[++l]='#'; printf("%d\n",manacher()); } return 0; }