【CF906E】Reverses（回文自动机，最小回文分割）

题意：给定两个长度相等的仅由小写字母组成的串A和B，问在A中最少选择多少段互不相交的子串进行翻转能使A和B相同

len<=5e5

思路：构造新串S=a[1]b[1]a[2]b[2]...a[n]b[n]

问题等价于求S的最小回文分割，其中需要每一段的长度都为偶数，注意长度为2的相当于没有翻转

把板子稍加修改即可

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 typedef pair<ll,int>P;
 #define N  1100010
 #define M  210000
 #define fi first
 #define se second
 #define MP make_pair
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const int MOD=,inv2=(MOD+)/;
       double eps=1e-;
       int INF=<<;
       ll inf=5e13;
       int dx[]={-,,,};
       int dy[]={,,-,};

 char a[N],b[N];
 int s[N];
 int n;
 int q,p,id,num[N],F[N],f[N],_f[N],pre[N],len[N],sk[N],df[N],t[N][];

 struct pam
 {
     void add(int x,int n)
     {
             while(s[n-len[p]-]!=s[n]) p=F[p];
         if(!t[p][x])
         {
                 int q=++id,k=F[p];
             len[q]=len[p]+;
             while(s[n-len[k]-]!=s[n]) k=F[k];
             F[q]=t[k][x];
             t[p][x]=q;
             df[q]=len[q]-len[F[q]];
             sk[q]=(df[q]==df[F[q]]?sk[F[q]]:F[q]);
         }
         p=t[p][x];
     }
 }pam;

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     scanf("%s",a+);
     int m=strlen(a+);
     scanf("%s",b+);
     int n=;
     rep(i,,m)
     {
         s[++n]=a[i]-'a';
         s[++n]=b[i]-'a';
     }
     s[n+]=s[]=;
     id=F[]=; len[]=-; _f[]=;
     rep(i,,n) f[i]=1e9;
     rep(i,,n)
     {
         pam.add(s[i],i);
         for(int x=p;x;x=sk[x])
         {
             _f[x]=i-len[sk[x]]-df[x];
             if(df[F[x]]==df[x]&&f[_f[x]]>f[_f[F[x]]]) _f[x]=_f[F[x]];
             if(i%==&&f[i]>f[_f[x]]+) f[i]=f[_f[x]]+,pre[i]=_f[x];
             if(i%==&&s[i]==s[i-]&&f[i-]<f[i])
             {
                 f[i]=f[i-];
                 pre[i]=i-;
             }
         }
     }
     //rep(i,1,n) printf("%d ",s[i]);
     //printf("\n");
     if(f[n]==1e9)
     {
         printf("-1\n");
         return ;
     }
     printf("%d\n",f[n]);
     int k=n;
     while(k)
     {
         int t=pre[k];
         if(t<k-) printf("%d %d\n",t/+,k/);
         k=t;
     }

     return ;
 }