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 = 1000020;

char str[maxn],solve[maxn<<1];
int f[maxn<<1],n;

void manacher(){
    memset(f,0,sizeof(f));
    for(int i=2*n-2,j=n-1;i>=2;i-=2,j--){
    solve[i] = str[j];solve[i-1] = '$';
    }
    solve[0] = str[0];
    f[0] = 1; n = 2*n-1;
    int fr = 0,last = 0;
    for(int i=1;i<n;i++){
    if(last + fr >= i){
        if(last-(i-last)-f[last-(i-last)]+1 >= last- fr)f[i]=min(n-i,f[last-(i-last)]);
        else f[i] = min(n-i,last+fr-i+1);
        while(i-f[i]>=0&&f[i]+i<n&&solve[f[i]+i]==solve[i-f[i]])
        f[i]++;
        if(i+f[i]-1 >= last+fr) last = i,fr = f[i]-1;
    }else{
        f[i] = 1;
        while(i-f[i]>=0&&f[i]+i<n&&solve[f[i]+i]==solve[i-f[i]]) f[i]++;
        last = i; fr = f[i]-1;
    }
    }
}

int ans[maxn];
void work(){
    memset(ans,0,sizeof(ans));
    n = strlen(str);int trans = n;
    manacher(); n = trans;
    for(int i=n-1;i>=0;i--){
    int len =  (f[i*2]+1)/2;
    if(i + len == n) {ans[i] = 1;continue;}
    if(i - len + 1 == 0){ans[i] = ans[i+len-1];continue;}
    }
    for(int i=0;i<n;i++) if(ans[i]) printf("%d ",i+1);
    puts("");
}

int main(){
    int t; scanf("%d",&t);
    while(t--){
    scanf("%s",str);
    work();
    }
    return 0;
}