LOJ6387 [THUPC2018] 绿绿与串串 【manacher】

题目分析:

比较简单，先跑一边manacher,然后对于回文部分可以碰到末尾的一定满足条件，否则向后转移。

代码:

 #include<bits/stdc++.h>
 using namespace std;

 const int maxn = ;

 char str[maxn],solve[maxn<<];
 int f[maxn<<],n;

 void manacher(){
     memset(f,,sizeof(f));
     for(int i=*n-,j=n-;i>=;i-=,j--){
     solve[i] = str[j];solve[i-] = '$';
     }
     solve[] = str[];
     f[] = ; n = *n-;
     int fr = ,last = ;
     for(int i=;i<n;i++){
     if(last + fr >= i){
         if(last-(i-last)-f[last-(i-last)]+ >= last- fr)f[i]=min(n-i,f[last-(i-last)]);
         else f[i] = min(n-i,last+fr-i+);
         while(i-f[i]>=&&f[i]+i<n&&solve[f[i]+i]==solve[i-f[i]])
         f[i]++;
         if(i+f[i]- >= last+fr) last = i,fr = f[i]-;
     }else{
         f[i] = ;
         while(i-f[i]>=&&f[i]+i<n&&solve[f[i]+i]==solve[i-f[i]]) f[i]++;
         last = i; fr = f[i]-;
     }
     }
 }

 int ans[maxn];
 void work(){
     memset(ans,,sizeof(ans));
     n = strlen(str);int trans = n;
     manacher(); n = trans;
     for(int i=n-;i>=;i--){
     int len =  (f[i*]+)/;
     if(i + len == n) {ans[i] = ;continue;}
     if(i - len +  == ){ans[i] = ans[i+len-];continue;}
     }
     for(int i=;i<n;i++) if(ans[i]) printf("%d ",i+);
     puts("");
 }

 int main(){
     int t; scanf("%d",&t);
     while(t--){
     scanf("%s",str);
     work();
     }
     return ;
 }