【BZOJ3238】【AHOI2013】差异

问了lcp了显然要后缀数组对吧

然后就变成了听说是HNOI2016 D2T1的简化版？单调栈处理出每个点向左右扩展的距离（贡献范围）即可。

// <3238.cpp> - 08/16/16 19:57:48
// This file is created by XuYike's black technology automatically.
// Copyright (C) 2015 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long lol;
int gi(){
    int res=0,fh=1;char ch=getchar();
    while((ch>'9'||ch<'0')&&ch!='-')ch=getchar();
    if(ch=='-')fh=-1,ch=getchar();
    while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();
    return fh*res;
}
const int MAXN=500010;
const int INF=1e9;
char s[MAXN];
int n;
int sa[MAXN],rk[MAXN];
int v[MAXN],x[MAXN],y[MAXN];
int h[MAXN];
void go(){
    int m=29;
    for(int i=1;i<=n;i++)v[x[i]=(s[i]-'a'+1)]++;
    for(int i=2;i<=m;i++)v[i]+=v[i-1];
    for(int i=n;i;i--)sa[v[x[i]]--]=i;
    for(int j=1;j<=n;j<<=1){
        int p=0;
        for(int i=n-j+1;i<=n;i++)y[++p]=i;
        for(int i=1;i<=n;i++)if(sa[i]>j)y[++p]=sa[i]-j;
        for(int i=1;i<=m;i++)v[i]=0;
        for(int i=1;i<=n;i++)v[x[y[i]]]++;
        for(int i=2;i<=m;i++)v[i]+=v[i-1];
        for(int i=n;i;i--)sa[v[x[y[i]]]--]=y[i];
        swap(x,y);x[sa[1]]=1;
        p=1;
        for(int i=2;i<=n;i++)x[sa[i]]=(y[sa[i]]==y[sa[i-1]]&&y[sa[i]+j]==y[sa[i-1]+j])?p:++p;
        if(p==n)break;m=p;
    }
}
void calh(){
    int k=0,j;
    for(int i=1;i<=n;i++)rk[sa[i]]=i;
    for(int i=1;i<=n;h[rk[i++]]=k)
        for(k?k--:0,j=sa[rk[i]-1];s[i+k]==s[j+k];k++);
}
int st[MAXN],l[MAXN],r[MAXN];
int main(){
    scanf("%s",s+1);
    n=strlen(s+1);
    go();calh();
    int t=1;st[1]=1;
    for(int i=2;i<=n;i++){
        while(t>1&&h[i]<h[st[t]])t--;
        l[i]=st[t];
        st[++t]=i;
    }
    st[t=1]=n+1;
    for(int i=n;i>=2;i--){
        while(t>1&&h[i]<=h[st[t]])t--;
        r[i]=st[t];
        st[++t]=i;
    }
    lol ans=1ll*(n-1)*n*(n+1)>>1;
    for(int i=2;i<=n;i++)ans-=1ll*(i-l[i])*(r[i]-i)*h[i]<<1;
    printf("%lld",ans);
    return 0;
}

// <3238.cpp> - 08/16/16 19:57:48

// This file is created by XuYike's black technology automatically.

// I don't know what this program is.

#include <iostream>

#include <vector>

#include <algorithm>

#include <cstring>

#include <cstdio>

#include <cmath>

using namespace std;

typedef long long lol;

int gi(){

int res=0,fh=1;char ch=getchar();

while((ch>'9'||ch<'0')&&ch!='-')ch=getchar();

if(ch=='-')fh=-1,ch=getchar();

while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar();

return fh*res;

}

const int MAXN=500010;

const int INF=1e9;

char s[MAXN];

int n;

int sa[MAXN],rk[MAXN];

int v[MAXN],x[MAXN],y[MAXN];

int h[MAXN];

void go(){

int m=29;

for(int i=1;i<=n;i++)v[x[i]=(s[i]-'a'+1)]++;

for(int i=2;i<=m;i++)v[i]+=v[i-1];

for(int i=n;i;i--)sa[v[x[i]]--]=i;

for(int j=1;j<=n;j<<=1){

int p=0;

for(int i=n-j+1;i<=n;i++)y[++p]=i;

for(int i=1;i<=n;i++)if(sa[i]>j)y[++p]=sa[i]-j;

for(int i=1;i<=m;i++)v[i]=0;

for(int i=1;i<=n;i++)v[x[y[i]]]++;

for(int i=2;i<=m;i++)v[i]+=v[i-1];

for(int i=n;i;i--)sa[v[x[y[i]]]--]=y[i];

swap(x,y);x[sa[1]]=1;

p=1;

for(int i=2;i<=n;i++)x[sa[i]]=(y[sa[i]]==y[sa[i-1]]&&y[sa[i]+j]==y[sa[i-1]+j])?p:++p;

if(p==n)break;m=p;

}

void calh(){

int k=0,j;

for(int i=1;i<=n;i++)rk[sa[i]]=i;

for(int i=1;i<=n;h[rk[i++]]=k)

for(k?k--:0,j=sa[rk[i]-1];s[i+k]==s[j+k];k++);

}

int st[MAXN],l[MAXN],r[MAXN];

int main(){

scanf("%s",s+1);

n=strlen(s+1);

go();calh();

int t=1;st[1]=1;

for(int i=2;i<=n;i++){

while(t>1&&h[i]<h[st[t]])t--;

l[i]=st[t];

st[++t]=i;

}

st[t=1]=n+1;

for(int i=n;i>=2;i--){

while(t>1&&h[i]<=h[st[t]])t--;

r[i]=st[t];

st[++t]=i;

}

lol ans=1ll*(n-1)*n*(n+1)>>1;

for(int i=2;i<=n;i++)ans-=1ll*(i-l[i])*(r[i]-i)*h[i]<<1;

printf("%lld",ans);

return 0;

}