Now that you have proposed a fake post for the HC2 Facebook page, Heidi wants to measure the quality of the post before actually posting it. She recently came across a (possibly fake) article about the impact of fractal structure on multimedia messages and she is now trying to measure the self-similarity of the message, which is defined as
where the sum is over all nonempty strings p and 
is the number of occurences of p in s as a substring. (Note that the sum is infinite, but it only has a finite number of nonzero summands.)
Heidi refuses to do anything else until she knows how to calculate this self-similarity. Could you please help her? (If you would like to instead convince Heidi that a finite string cannot be a fractal anyway – do not bother, we have already tried.)
The input starts with a line indicating the number of test cases T (1 ≤ T ≤ 10). After that, T test cases follow, each of which consists of one line containing a string s (1 ≤ |s| ≤ 100 000) composed of lowercase letters (a-z).
Output T lines, every line containing one number – the answer to the corresponding test case.
4
aa
abcd
ccc
abcc
5
10
14
12
A string s contains another string p as a substring if p is a contiguous subsequence of s. For example, ab is a substring of cab but not of acb.
题目大意：求一个字符串所有子串的出现次数平方和
惊了，我会Hard但是不会做normal
这个，参考一下弦论的话就是SAM裸题吧
SAM的每个状态代表了一些子串，它们的出现次数是就是R集合的大小，而子串的个数就是maxl-minl+1
求出SAM，求出每个状态的R集合大小s，以及maxl和minl
最终的答案就是 $ sum (maxl-minl+1)s^2 $
R集合不是要拓扑排序么，事实上直接按照MAXL基数排序即可，从黄学长那学来的奇技淫巧
1 #include <iostream> 2 #include <cstdlib> 3 #include <cstdio> 4 #include <algorithm> 5 #include <string> 6 #include <cstring> 7 #include <cmath> 8 #include <map> 9 #include <stack> 10 #include <set> 11 #include <vector> 12 #include <queue> 13 #include <time.h> 14 #define eps 1e-7 15 #define INF 0x3f3f3f3f 16 #define MOD 1000000007 17 #define rep0(j,n) for(int j=0;j<n;++j) 18 #define rep1(j,n) for(int j=1;j<=n;++j) 19 #define pb push_back 20 #define set0(n) memset(n,0,sizeof(n)) 21 #define ll long long 22 #define ull unsigned long long 23 #define iter(i,v) for(edge *i=head[v];i;i=i->nxt) 24 #define max(a,b) (a>b?a:b) 25 #define min(a,b) (a<b?a:b) 26 #define print_runtime printf("Running time:%.3lfs ",double(clock())/1000.0) 27 #define TO(j) printf(#j": %d ",j); 28 //#define OJ 29 using namespace std; 30 const int MAXINT = 100010; 31 const int MAXNODE = 100010; 32 const int MAXEDGE = 2*MAXNODE; 33 char BUF,*buf; 34 int read(){ 35 char c=getchar();int f=1,x=0; 36 while(!isdigit(c)){if(c=='-') f=-1;c=getchar();} 37 while(isdigit(c)){x=x*10+c-'0';c=getchar();} 38 return f*x; 39 } 40 char get_ch(){ 41 char c=getchar(); 42 while(!isalpha(c)) c=getchar(); 43 return c; 44 } 45 //------------------- Head Files ----------------------// 46 char s[MAXINT]; 47 struct cond{ 48 cond *c[26],*fa; 49 int maxl,r,cnt; 50 }; 51 struct SAM{ 52 cond *root,*last; 53 cond mp[MAXINT*5]; 54 int ct[MAXINT],srt[MAXINT*5]; 55 int cnt; 56 cond * newnode(int maxl){ 57 cond *p = &mp[cnt++]; 58 memset(p->c,0,sizeof(p->c)); 59 p->fa=NULL; 60 p->maxl=maxl; 61 p->cnt=p->r=0; 62 return p; 63 } 64 SAM(){ 65 cnt=0; 66 root = newnode(0); 67 last=root; 68 } 69 void append(int w){ 70 cond *p=last; 71 cond *np=newnode(p->maxl+1); 72 np->r=1; 73 while(p&&!p->c[w]) p->c[w]=np,p=p->fa; 74 if(!p) np->fa=root; 75 else{ 76 cond *q=p->c[w]; 77 if(q->maxl==p->maxl+1) np->fa=q; 78 else{ 79 cond *nq = newnode(p->maxl+1); 80 memcpy(nq->c,q->c,sizeof(q->c)); 81 nq->fa=q->fa; 82 q->fa=nq;np->fa=nq; 83 while(p&&p->c[w]==q) p->c[w]=nq,p=p->fa; 84 } 85 } 86 last=np; 87 } 88 ll calc(){ 89 memset(ct,0,sizeof(ct)); 90 ll ans=0; 91 rep1(i,cnt-1) ct[mp[i].maxl]++; 92 rep1(i,MAXINT-1) ct[i]+=ct[i-1]; 93 rep1(i,cnt-1) srt[ct[mp[i].maxl]--]=i; 94 for(int i=cnt-1;i>=1;i--) { 95 mp[srt[i]].cnt+=mp[srt[i]].r; 96 mp[srt[i]].fa->cnt+=mp[srt[i]].cnt; 97 } 98 rep1(i,cnt-1) ans+=(ll)(mp[i].maxl-mp[i].fa->maxl)*mp[i].cnt*mp[i].cnt; 99 cnt=0; 100 root = newnode(0); 101 last=root; 102 return ans; 103 } 104 }sam; 105 void get_input(); 106 void work(); 107 int main() { 108 int T=read(); 109 while(T--){ 110 get_input(); 111 work(); 112 } 113 return 0; 114 } 115 void work(){ 116 printf("%lld ",sam.calc()); 117 } 118 void get_input(){ 119 scanf("%s",s); 120 int l =strlen(s); 121 rep0(i,l) sam.append(s[i]-'a'); 122 }