bzoj3277

题解：

后缀自动机

然后抄了一发题解

可以看看这个博客：http://blog.csdn.net/clover_hxy/article/details/53861268

代码：

#include<bits/stdc++.h>
using namespace std;
const int N=200005;
typedef long long ll;
int n,m,np,nq,p,q,cnt,root,last,len,sz,ch[N][20],fa[N];
int size[N],l[N],nxt[N],a[N],now,pd[N],mark[N],v[N],pos[N];  
int point[N],next[N],vi[N],tot,tt,head[N],nt[N];  
ll c[N];
char s[N];
void add(int x,int y)
{
    next[++tot]=point[x];
    point[x]=tot;
    vi[tot]=y;
}
void addfa(int x,int y)
{
    nt[++tt]=head[x];
    head[x]=tt;
    v[tt]=y;
}
void extend(int x)
{
    int c=a[x];
    p=last;
    np=++cnt;
    last=np;
    l[np]=l[p]+1;
    for (;p&&!ch[p][c];p=fa[p])ch[p][c]=np;
    if (!p)fa[np]=root;
    else
     {
         q=ch[p][c];
         if (l[q]==l[p]+1)fa[np]=q;
         else
          {
              nq=++cnt;
              l[nq]=l[p]+1;
              memcpy(ch[nq],ch[q],sizeof ch[nq]);
              size[nq]=size[q];
              nxt[nq]=nxt[q];
              fa[nq]=fa[q];
              fa[q]=fa[np]=nq;
              for (;ch[p][c]==q;p=fa[p])ch[p][c]=nq;
         }
     }
    add(now,np);
    for (;np;np=fa[np])
     if (nxt[np]!=now)
      {
          nxt[np]=now;
          size[np]++;
      } 
     else break;  
}
void dfs(int x)
{
    c[x]+=c[fa[x]];
    for (int i=head[x];i;i=nt[i])dfs(v[i]);
}
int main()
{
    scanf("%d%d",&n,&m);
    root=++cnt;
    for (int i=1;i<=n;i++)
     {
         scanf("%s",s+1);
         last=root;
         len=strlen(s+1);
         now=i;
         for (int j=1;j<=len;j++)
          a[++sz]=s[j]-'a',extend(sz);
         mark[i]=sz;
     }
    for (int i=1;i<=cnt;i++)v[l[i]]++;
    for (int i=1;i<=cnt;i++)v[i]+=v[i-1];
    for (int i=1;i<=cnt;i++)pos[v[l[i]]--]=i;
    for (int i=1;i<=cnt;i++)
     {
         int t=pos[i];
         addfa(fa[t],t);
         if (size[t]>=m)c[t]=(ll)(l[t]-l[fa[t]]);
         else c[t]=0;
     } 
    c[1]=0;
    dfs(1);
    for (int i=1;i<=n;i++)
     {
         ll ans=0;
         for (int j=point[i];j;j=next[j])ans+=c[vi[j]];
         printf("%lld ",ans);
     }
}