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  • 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


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