并不对劲的bzoj4650:loj2083:uoj219:p1117:[NOI2016]优秀的拆分

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
*/