hdu_3294_Girls' research(Manacher)

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=3294

题意:给你一个字符和一个字符串,第一个字符表示该字符代表a,然后让你求变换后的最长回文区间并输出(这里的回文长度要大于1)

题解:直接上马拉车,然后记录一下区间

 1 #include<cstdio>
 2 #include<cstring>
 3 #define min(a,b) (a)>(b)?(b):(a)
 4 #define max(a,b) (a)>(b)?(a):(b)
 5 const int maxn = 200005;//字符串长度
 6 int lb,rb;//左右区间
 7 struct Manacher{
 8     char str[maxn<<1];
 9     int p[maxn<<1],len,mx,id,tl,ans,i;
10     int maxlen(char *s){
11         len=strlen(s),mx=0,id=0,tl=0,str[tl++]='$',str[tl++]='#';
12         for(i=0;i<len;i++)str[tl++]=s[i],str[tl++]='#';
13         for(i=2,str[tl]=0,ans=0;i<tl;i++){
14             p[i]=mx>i?min(p[(id<<1)-i],mx-i):1;
15             while(str[i-p[i]]==str[i+p[i]])p[i]++;
16             if(i+p[i]>mx)mx=i+p[i],id=i;
17             if(ans<p[i])ans=p[i],lb=(i-p[i])/2,rb=(i+p[i])/2-2;
18         }
19         return ans-1;
20     }
21 }M;
22 char s[maxn],tt[2];
23 int main(){
24     while(~scanf("%s%s",tt,s)){
25         for(int i=0,tmp=tt[0]-'a';s[i]!='';i++){
26             s[i]-=tmp;
27             if(s[i]<'a')s[i]=s[i]-'a'+'z'+1;
28             }
29         int r=M.maxlen(s);
30         if(r>1){
31             printf("%d %d
",lb,rb);
32             for(int i=lb;i<=rb;i++)printf("%c",s[i]);
33             printf("
");
34         }else printf("No solution!
");
35     }
36     return 0;
37 }
View Code


原文地址:https://www.cnblogs.com/bin-gege/p/5696178.html