bzoj 3238

明显要用求SA再求h，而lcp(i,j)就等于i到j之间最小的h

用f[i]表示i跟比i大的后缀的lcp的和，然后用单调栈维护最小值就好了（有点像DP？）

 1 #include<bits/stdc++.h>
 2 #define inc(i,l,r) for(int i=l;i<=r;i++)
 3 #define dec(i,l,r) for(int i=l;i>=r;i--)
 4 #define link(x) for(edge *j=h[x];j;j=j->next)
 5 #define mem(a) memset(a,0,sizeof(a))
 6 #define inf 1e9
 7 #define ll long long
 8 #define succ(x) (1<<x)
 9 #define NM 500000+5
10 using namespace std;
11 int read(){
12     int x=0,f=1;char ch=getchar();
13     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
14     while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
15     return x*f;
16 }
17 char st[NM];
18 int n,m,tmp[NM],top[NM],rank[NM],sa[NM],h[NM];
19 ll f[NM],ans;
20 stack<int >s;
21 void getsa(){
22     int j;m=256;
23     inc(i,0,n)top[rank[i]=(int)st[i]]++;
24     inc(i,1,m)top[i]+=top[i-1];
25     inc(i,0,n)sa[--top[rank[i]]]=i;
26 //    inc(i,0,n)printf("%d ",sa[i]);printf("\n");
27     for(int k=1;k<=n;k<<=1){
28         inc(i,0,n){
29             j=sa[i]-k;
30             if(j<0)j+=n+1;
31             tmp[top[rank[j]]++]=j;
32         }
33 //        inc(i,0,n)printf("%d ",tmp[i]);printf("\n");
34         sa[tmp[top[0]=0]]=j=0;
35         inc(i,1,n){
36             if(rank[tmp[i-1]]!=rank[tmp[i]]||rank[tmp[i]+k]!=rank[tmp[i-1]+k])
37             top[++j]=i;
38             sa[tmp[i]]=j;
39         }
40         memcpy(rank,sa,sizeof(sa));
41         memcpy(sa,tmp,sizeof(tmp));
42         if(j>=n)break;
43     }
44     j=0;
45 //    inc(i,0,n)printf("%d ",sa[i]);printf("\n");
46     inc(i,0,n)if(rank[i]){
47         if(j)j--;
48         while(st[i+j]==st[sa[rank[i]-1]+j])
49         j++;
50         h[rank[i]]=j;
51     }
52 //    inc(i,0,n)printf("%d ",h[i]);printf("\n");
53 }
54 int main(){
55     freopen("data.in","r",stdin);
56     scanf("%s",st);
57     n=strlen(st);st[n+1]='$';
58     getsa();
59     ans=(ll)(n+1)*(n-1)*n>>1;
60     dec(i,n,0){
61         while(!s.empty()&&h[s.top()]>h[i])s.pop();
62         int t=s.empty()?n+1:s.top();
63         f[i]=(ll)h[i]*(t-i)+f[t];
64         ans-=f[i]<<1;s.push(i);
65     }
66 //    inc(i,0,n)printf("%lld ",f[i]);printf("\n");
67     printf("%lld",ans);
68     return 0;
69 }

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