题目大意
“优秀的拆分”指将一个字符串拆分成AABB的形式
十次询问，每次给出一个字符串S（\(|S|\leq3*10^4\)），求它的所有子串的优秀的拆分的方案数之和
题解
此题过于优秀，题解先坑着
代码
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
#define view(u,k) for(int k=fir[u];k!=-1;k=nxt[k])
#define maxn 30010
#define LL long long
using namespace std;
int read()
{
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)&&ch!='-')ch=getchar();
if(ch=='-')f=-1,ch=getchar();
while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
return x*f;
}
void write(LL x)
{
if(x==0){putchar('0'),putchar('\n');return;}
int f=0;char ch[20];
if(x<0)putchar('-'),x=-x;
while(x)ch[++f]=x%10+'0',x/=10;
while(f)putchar(ch[f--]);
putchar('\n');
return;
}
int t,n,a[2][maxn];
LL ans;
char s[maxn];
void add(int l,int r,int f){if(l<=r)a[f][l]++,a[f][r+1]--;}
void geta(){rep(i,1,n)a[0][i]+=a[0][i-1],a[1][i]+=a[1][i-1];return;}
struct SAM
{
int ch[maxn<<1][26],rt,lst,cnt,dis[maxn<<1],fa[maxn<<1],pos[maxn],tim,dfn[maxn<<1],st[20][maxn<<2],lg[maxn<<2],len;
int fir[maxn<<1],nxt[maxn<<1],v[maxn<<1],cnte,dep[maxn<<1];
char s[maxn];
void ade(int u1,int v1){v[cnte]=v1,nxt[cnte]=fir[u1],fir[u1]=cnte++;}
int gx(char c){return c-'a';}
void ext(int id)
{
int p=lst,np=++cnt,val=gx(s[id]);dis[np]=id,pos[id]=np,lst=np;
for(;p&&!ch[p][val];p=fa[p])ch[p][val]=np;
if(!p)fa[np]=rt;
else
{
int q=ch[p][val];
if(dis[q]==dis[p]+1)fa[np]=q;
else
{
int nq=++cnt;dis[nq]=dis[p]+1;
fa[nq]=fa[q],fa[q]=fa[np]=nq;
memcpy(ch[nq],ch[q],sizeof(ch[q]));
for(;p&&ch[p][val]==q;p=fa[p])ch[p][val]=nq;
}
}
}
void dfs(int u)
{
dfn[u]=++tim,st[0][tim]=u;
view(u,k){dep[v[k]]=dep[u]+1,dfs(v[k]),st[0][++tim]=u;}
}
int lcs(int x,int y)
{
x=dfn[pos[x]],y=dfn[pos[y]];
if(x>y)swap(x,y);
int len=y-x+1;
return dis[dep[st[lg[len]][x]]<dep[st[lg[len]][y-(1<<lg[len])+1]]?st[lg[len]][x]:st[lg[len]][y-(1<<lg[len])+1]];
}
void build()
{
rt=lst=cnt=1;
rep(i,1,len)ext(i);
rep(i,1,cnt)fir[i]=-1;
rep(i,1,cnt)ade(fa[i],i);lg[0]=-1;
dfs(rt);
rep(i,1,tim)lg[i]=lg[i>>1]+1;
rep(k,1,lg[tim])for(int i=1;i+(1<<k)-1<=tim;i++)
st[k][i]=dep[st[k-1][i]]<dep[st[k-1][i+(1<<(k-1))]]?st[k-1][i]:st[k-1][i+(1<<(k-1))];
}
void reset()
{
rep(i,1,cnt){dep[i]=fa[i]=dis[i]=dfn[i]=0;rep(j,0,25)ch[i][j]=0;}
rep(i,1,tim)st[0][i]=0;tim=cnt=rt=cnte=lst=0;
}
}pre,suf;
int main()
{
t=read();
while(t--)
{
scanf("%s",s+1);
n=strlen(s+1);ans=0;
rep(i,1,n)pre.s[i]=s[i],suf.s[i]=s[n-i+1];pre.len=suf.len=n;
pre.build(),suf.build();
rep(len,1,(n>>1))
{
for(int i=1;i+len<=n;i+=len)
{
int lp=pre.lcs(i,i+len),ls=suf.lcs(n-i+1,n-(i+len)+1);
if(lp+ls>len)add(max(i-len+1,i-lp+1),min(i,i+ls-len),1),add(max(i+len,i-lp+(len<<1)),min(i+len+len-1,i+len+ls-1),0);
}
}
geta();
rep(i,1,n-1)ans+=(LL)a[0][i]*(LL)a[1][i+1];
rep(i,0,n+1)a[0][i]=a[1][i]=0;pre.reset(),suf.reset();
write(ans);
}
return 0;
}
/*
4
aabbbb
cccccc
aabaabaabaa
bbaabaababaaba
*/