bzoj3230 相似子串

题目链接

每个子串都是一个后缀的前缀

每个后缀贡献的子串数目是len-sa[i]-height[i];

因此可以二分找到一个子串的位置，要求某两个子串的最长公共前缀和最长公共后缀，把原串倒过再来一发就好，然后st表O(1)查询;

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define rre(i,r,l) for(int i=(r);i>=(l);i--)
#define re(i,l,r) for(int i=(l);i<=(r);i++)
#define Clear(a,b) memset(a,b,sizeof(a))
#define inout(x) printf("%d",(x))
#define douin(x) scanf("%lf",&x)
#define strin(x) scanf("%s",(x))
#define LLin(x) scanf("%lld",&x)
#define op operator
#define CSC main
typedef unsigned long long ULL;
typedef const int cint;
typedef long long LL;
using namespace std;
void inin(int &ret)
{
    ret=0;int f=0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=1;ch=getchar();}
    while(ch>='0'&&ch<='9')ret*=10,ret+=ch-'0',ch=getchar();
    ret=f?-ret:ret;
}
char s[100010];
int LOG[100010],n,q;
cint inf=2147483647;
struct string
{
    int sa[100010],h[100010],t[100010],t2[100010],c[100010],rank[100010];
    LL sum[100010];
    int f[100010][20];
    void build_sa(int m)
    {
        int *x=t,*y=t2;
        re(i,0,m-1)c[i]=0;
        re(i,0,n-1)x[i]=s[i],c[x[i]]++;
        re(i,1,m-1)c[i]+=c[i-1];
        rre(i,n-1,0)sa[--c[x[i]]]=i;
        for(int k=1;k<=n;k<<=1)
        {
            int p=0;
            rre(i,n-1,n-k)y[p++]=i;
            re(i,0,n-1)if(sa[i]>=k)y[p++]=sa[i]-k;
            re(i,0,m-1)c[i]=0;
            re(i,0,n-1)c[x[y[i]]]++;
            re(i,1,m-1)c[i]+=c[i-1];
            rre(i,n-1,0)sa[--c[x[y[i]]]]=y[i];
            swap(x,y);
            x[sa[0]]=0;p=1;
            re(i,1,n-1)x[sa[i]]=y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k]?p-1:p++;
            if(p>=n)return ;
            m=p;
        }
    }
    void build_height()//
    {
        int k=0;
        re(i,0,n-1)rank[sa[i]]=i;
        re(i,0,n-1)
        {
            if(k)k--;
            if(!rank[i])continue ;
            int j=sa[rank[i]-1];
            while(s[i+k]==s[j+k])k++;
            h[rank[i]]=k;
        }
    }
    void build_sum()//
    {
        sum[0]=n-sa[0];
        re(i,1,n-1)sum[i]=sum[i-1]+n-h[i]-sa[i];
    }
    void build_RMQ()//
    {
        re(i,1,n-1)f[i][0]=h[i];
        for(int j=1;(1<<j)<n;j++)
            for(int i=1;i+(1<<(j-1))<n;i++)
                f[i][j]=min(f[i][j-1],f[i+(1<<(j-1))][j-1]);
    }
    void init()
    {
        build_sa(256);
        build_height();
        build_sum();
        build_RMQ();
    }
    void find(LL wei,int &a,int &b)//
    {
        int x=lower_bound(sum,sum+n,wei)-sum;
        a=sa[x],b=sa[x]+h[x]+wei-(x?sum[x-1]:0);
    }
    int query(int l,int r)//
    {
        l=rank[l],r=rank[r];
        if(r<l)swap(l,r);l++;
        int k=LOG[r-l+1];
        return min(f[l][k],f[r-(1<<k)+1][k]);
    }
}A,B;
int main()
{
    freopen("in.in","r",stdin);
    freopen("out.out","w",stdout);
    inin(n),inin(q);
    LOG[0]=-1;
    re(i,1,100000)LOG[i]=LOG[i>>1]+1;
    strin(s);
    A.init();
    reverse(s,s+n);
    B.init();
    while(q--)
    {
        int a1,b1,a2,b2,t;
        LL l,r,ans=0;
        LLin(l),LLin(r);
        if(l>A.sum[n-1]||r>A.sum[n-1])
        {
            cout<<"-1
";
            continue;
        }
        A.find(l,a1,b1);
        A.find(r,a2,b2);
        t=a1==a2?inf:A.query(a1,a2);
        t=min(t,min(b1-a1,b2-a2));
        ans+=(LL)t*t;
        t=b1==b2?inf:B.query(n-b1,n-b2);
        t=min(t,min(b1-a1,b2-a2));
        ans+=(LL)t*t;
        printf("%lld
",ans);
    }
     return 0;
}