传送门->
又称普及大会。
这题没什么好说的……后缀自动机裸题……并不对劲的人太菜了，之前照着标程逐行比对才过了这道题，前几天刚刚把这题一遍写对……
这题的输出和某两点相同后缀的长度有关，那么把串反过来就和相同前缀的长度有关。建出后缀自动机后，发现点u代表了right[u]个dis[fa[u]+1]~dis[u]相似的子串。此时如何统计答案就很显然了。
想着很简单，写着嘛…其实并不长？
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register LL i=(x);i<=(y);i++)
#define dwn(i,x,y) for(register LL i=(x);i>=(y);i--)
#define re register
#define maxn 600010
#define di ord[i]
#define LL long long
using namespace std;
inline LL read()
{
LL x=0,f=1;
char ch=getchar();
while(isdigit(ch)==0 && ch!='-')ch=getchar();
if(ch=='-')f=-1,ch=getchar();
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return x*f;
}
inline void write_(LL x)
{
LL f=0;char ch[20];
if(!x){putchar('0'),putchar(' ');return;}
if(x<0){putchar('-');x=-x;}
while(x)ch[++f]=x%10+'0',x/=10;
while(f)putchar(ch[f--]);
putchar(' ');
}
inline void writen(LL x)
{
LL f=0;char ch[20];
if(!x){puts("0");return;}
if(x<0){putchar('-');x=-x;}
while(x)ch[++f]=x%10+'0',x/=10;
while(f)putchar(ch[f--]);
putchar('
');
}
LL ch[maxn][30],fa[maxn],r[maxn],dis[maxn],mn[maxn],mns[maxn],mx[maxn],mxs[maxn],lst,cnt;
LL cc[maxn],ord[maxn],rt,n,a[maxn],ans[maxn],num[maxn],inf[4];
char s[maxn];
LL gx(char c){return c-'a';}
void extend(char c)
{
LL np=++cnt,p=lst;dis[np]=dis[lst]+1,lst=np;
for(;p&&!ch[p][gx(c)];p=fa[p])ch[p][gx(c)]=np;
if(p==0)fa[np]=rt;
else
{
LL q=ch[p][gx(c)];
if(dis[q]==dis[p]+1)fa[np]=q;
else
{
LL 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(;ch[p][gx(c)]==q;p=fa[p])ch[p][gx(c)]=nq;
}
}
}
void upd(LL x,LL y)
{
r[x]+=r[y];
if(mn[x]<mn[y])mns[x]=min(mns[x],mn[y]);
else mns[x]=min(mns[y],mn[x]),mn[x]=mn[y];
if(mx[x]>mx[y])mxs[x]=max(mxs[x],mx[y]);
else mxs[x]=max(mxs[y],mx[x]),mx[x]=mx[y];
}
LL mul(LL x)
{
LL tmx,tmn;
if(mx[x]!=-inf[0]&&mxs[x]!=-inf[0])tmx=mx[x]*mxs[x];
else tmx=-inf[0];
if(mn[x]!=inf[0]&&mns[x]!=inf[0])tmn=mn[x]*mns[x];
else tmn=-inf[0];
return max(tmx,tmn);
}
void qsort()
{
memset(cc,0,sizeof(cc));
rep(i,1,cnt)cc[dis[i]]++;
rep(i,1,n)cc[i]+=cc[i-1];
rep(i,1,cnt)ord[cc[dis[i]]--]=i;
}
int main()
{
n=read();
lst=rt=++cnt;dis[0]=-1;
scanf("%s",s+1);
rep(i,1,n)a[i]=read();
rep(i,1,(n>>1))swap(s[i],s[n-i+1]),swap(a[i],a[n-i+1]);
rep(i,1,n)extend(s[i]);
qsort();
memset(inf,0x7f,sizeof(inf));
rep(i,1,cnt)mx[i]=mxs[i]=ans[i]=-inf[0],mn[i]=mns[i]=inf[0];
LL p=rt;
rep(i,1,n)p=ch[p][gx(s[i])],mx[p]=mn[p]=a[i],r[p]=1;
dwn(i,cnt,1)upd(fa[di],di);
rep(i,1,cnt)
ans[dis[i]]=max(mul(i),ans[dis[i]]),
num[dis[fa[i]]+1]+=r[i]*(r[i]-1)/2,num[dis[i]+1]-=r[i]*(r[i]-1)/2;
dwn(i,n,1)ans[i]=max(ans[i],ans[i+1]);
rep(i,0,n-1)
num[i]+=num[i-1],write_(num[i]),
writen(num[i]==0?0:ans[i]);
return 0;
}
/*
10
ponoiiipoi
2 1 4 7 4 8 3 6 4 7
*/
据说用后缀数组也能做，不过思维难度略高。
要是觉得对后缀自动机还不够熟悉，可以试一下bzoj3238: [Ahoi2013]，也是后缀自动机很好写（才怪）的题。
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register LL i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register LL i=(x);i>=(y);--i)
#define re register
#define maxn 1000010
#define LL long long
using namespace std;
inline LL read()
{
LL x=0,f=1;
char ch=getchar();
while(isdigit(ch)==0 && ch!='-')ch=getchar();
if(ch=='-')f=-1,ch=getchar();
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return x*f;
}
inline void write(LL x)
{
LL f=0;char ch[20];
if(!x){puts("0");return;}
if(x<0){putchar('-');x=-x;}
while(x)ch[++f]=x%10+'0',x/=10;
while(f)putchar(ch[f--]);
putchar('
');
}
LL ch[maxn][30],fa[maxn],dis[maxn],yes[maxn],lst,rt,cntnd;
LL v[maxn],nxt[maxn],fir[maxn],siz[maxn],ans[maxn],cntrd,len;
char s[maxn];
void ade(LL u1,LL v1){v[++cntrd]=v1,nxt[cntrd]=fir[u1],fir[u1]=cntrd;}//u->v
LL gx(char c){return c-'a';}
void extend(char c)
{
LL np=++cntnd,p=lst;dis[np]=dis[p]+1,lst=np;
for(;p&&ch[p][gx(c)]==0;p=fa[p])ch[p][gx(c)]=np;
if(!p)fa[np]=rt;
else
{
LL q=ch[p][gx(c)];
if(dis[q]==dis[p]+1)fa[np]=q;
else
{
LL nq=++cntnd;dis[nq]=dis[p]+1;
fa[nq]=fa[q],fa[np]=fa[q]=nq;
memcpy(ch[nq],ch[q],sizeof(ch[q]));
for(;ch[p][gx(c)]==q;p=fa[p])ch[p][gx(c)]=nq;
}
}
}
void getans(LL u)
{
if(yes[u])siz[u]=1;
LL tmp=0;
for(LL k=fir[u];k!=-1;k=nxt[k])
{
getans(v[k]);
siz[u]+=siz[v[k]];
}
for(LL k=fir[u];k!=-1;k=nxt[k])
tmp+=siz[v[k]]*(siz[u]-siz[v[k]]);
ans[u]=tmp*dis[u]+yes[u]*(siz[u]-1ll)*dis[u];
}
int main()
{
memset(fir,-1,sizeof(fir));
memset(ans,0,sizeof(ans));
lst=rt=++cntnd;dis[0]=-1;
scanf("%s",s+1);
len=strlen(s+1);
rep(i,1,len)swap(s[i],s[len-i+1]);
rep(i,1,len)extend(s[i]);
LL p=rt;
rep(i,1,cntnd)ade(fa[i],i);
rep(i,1,len)p=ch[p][gx(s[i])],yes[p]=1;
getans(rt);
LL ansans=0;
rep(i,1,cntnd)ansans+=ans[i];
write((len+1ll)*len*(len-1ll)/2ll-ansans);
return 0;
}
/*
cacao
*/
最后祝您身体健康，再见。