bzoj3238 [Ahoi2013]差异

题目链接

树状数组+单调栈

计算出每个后缀的前面、后面第一个h[]比它小的(前闭后开)，乘起来计算答案

#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[500050];
int n,sa[500050],c[500050],t[500050],t2[500050],h[500050];
void build_sa(int m)
{
    int *x=t,*y=t2;
    re(i,0,n-1)x[i]=s[i],c[x[i]]++;
    re(i,1,m)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,0,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;
    }
}
int rank[500050];
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;
    }
}
int l[500050],r[500050],sta[500050];
LL ans;
int main()
{
    strin(s);
    n=strlen(s);
    build_sa(256);
    build_height();
    int top=0;
    re(i,0,n-1)
    {
        while(top&&h[sta[top]]>=h[i])top--;
        l[i]=(top?sta[top]+1:0);
        sta[++top]=i;
    }
    top=0;
    rre(i,n-1,0)
    {
        while(top&&h[sta[top]]>h[i])top--;
        r[i]=(top?sta[top]-1:n-1);
        sta[++top]=i;
    }
    ans+=(LL)(1+n)*n/2*(n-1);
    re(i,0,n-1)
        ans-=2LL*(i-l[i]+1)*(r[i]-i+1)*h[i];
    printf("%lld",ans);
     return 0;
}