[bzoj2565]最长双回文子串

题目大意

给定字符串求其最长双回文子串，双回文子串的定义是该串可以分成两个串它们都是回文串。

题解

详见GDKOI2016的项链
我直接超标了（不知所措

#include<cstdio>#include<cstring>#include<algorithm>#define fo(i,a,b) for(i=a;i<=b;i++)#define fd(i,a,b) for(i=a;i>=b;i--)using namespace std;const int maxn=100000+10;int f[maxn*4];int tree[maxn*20];int i,j,k,l,r,t,n,m,ans,tot;char ch;char s[maxn*2],h[maxn*4];void change(int p,int l,int r,int a,int b){    if (l==r){        tree[p]=b;        return;    }    int mid=(l+r)/2;    if (a<=mid) change(p*2,l,mid,a,b);else change(p*2+1,mid+1,r,a,b);    tree[p]=max(tree[p*2],tree[p*2+1]);}int query(int p,int l,int r,int a,int b){    if (a>b) return 0;    if (l==a&&r==b) return tree[p];    int mid=(l+r)/2;    if (b<=mid) return query(p*2,l,mid,a,b);    else if (a>mid) return query(p*2+1,mid+1,r,a,b);    else return max(query(p*2,l,mid,a,mid),query(p*2+1,mid+1,r,mid+1,b));}int main(){    while (1){        ch=getchar();        if (ch<'a'||ch>'z') break;        s[++n]=ch;    }    h[0]='@';    h[1]='$';    fo(i,1,n) h[i*2]=s[i],h[i*2+1]='$';    h[n*2+2]='%';    j=0;    f[0]=0;    fo(i,2,n*2){        if (i>j+f[j]){            f[i]=0;            while (h[i+f[i]+1]==h[i-f[i]-1]) f[i]++;        }        else{            f[i]=min(f[2*j-i],j+f[j]-i);            while (h[i+f[i]+1]==h[i-f[i]-1]) f[i]++;        }        if (i+f[i]>j+f[j]) j=i;    }    fo(i,3,n+2) change(1,1,n*2+1,i-f[i],i);    fo(i,2,n*2){        ans=max(ans,query(1,1,n*2+1,1,i+f[i])-i);        if (i+n+1<=n*2) change(1,1,n*2+1,i+n+1-f[i+n+1],i+n+1);    }    printf("%d\n",ans);}