Description
Just like humans, cows often appreciate feeling they are unique in some way. Since Farmer John's cow
s all come from the same breed and look quite similar, they want to measure uniqueness in their name
s.Each cow's name has some number of substrings. For example, "amy" has substrings {a, m, y, am, my,
amy}, and "tommy" would have the following substrings: {t, o, m, y, to, om, mm, my, tom, omm, mmy,
tomm, ommy, tommy}.A cow name has a "uniqueness factor" which is the number of substrings of that na
me not shared with any other cow. For example, If amy was in a herd by herself, her uniqueness facto
r would be 6. If tommy was in a herd by himself, his uniqueness factor would be 14. If they were in
a herd together, however, amy's uniqueness factor would be 3 and tommy's would be 11.Given a herd of
cows, please determine each cow's uniqueness factor.
定义一个字符串的「独特值」为只属于该字符串的本质不同的非空子串的个数。如 "amy" 与 “tommy” 两个串，
只属于 "amy" 的本质不同的子串为 "a" "am" "amy" 共 3 个。只属于 "tommy" 的本质不同的子串为 "t" "to" "
tom" "tomm" "tommy" "o" "om" "omm" "ommy" "mm" "mmy" 共 11 个。 所以 "amy" 的「独特值」为 3 ，"tommy
" 的「独特值」为 11 。给定 N 个字符集为小写英文字母的字符串，所有字符串的长度和小于 10^5 ，求出每个
字符串「独特值」
Input
The first line of input will contain NN (1≤N≤10^5).
The following NN lines will each contain the name of a cow in the herd.
Each name will contain only lowercase characters a-z.
The total length of all names will not exceed 10^5
Output
Output NN numbers, one per line, describing the uniqueness factor of each cow.
Sample Input
3
amy
tommy
bessie
amy
tommy
bessie
Sample Output
3
11
19
11
19
解题思路：
这次略有不同，建完Parent树之后，要在一个节点中统计子串种类。
但是并不需要想Dfs序那样，只需要统计是否大于1就好了。
代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 typedef long long lnt; 5 struct sant{ 6 int tranc[26]; 7 int len; 8 int pre; 9 }s[200000],sts; 10 struct pnt{ 11 int hd; 12 int wgt; 13 int blg; 14 }p[200000]; 15 struct ent{ 16 int twd; 17 int lst; 18 }e[200000]; 19 int fin; 20 int siz; 21 int n,T; 22 int cnt; 23 lnt ans[200000]; 24 char tmp[200000]; 25 void ade(int f,int t) 26 { 27 cnt++; 28 e[cnt].twd=t; 29 e[cnt].lst=p[f].hd; 30 p[f].hd=cnt; 31 return ; 32 } 33 void res(void) 34 { 35 fin=siz=1; 36 s[1]=s[0]=sts; 37 return ; 38 } 39 void Insert(int c,int i) 40 { 41 int nwp,lsp,nwq,lsq; 42 nwp=++siz; 43 p[nwp].blg=i; 44 p[nwp].wgt=1; 45 s[nwp].len=s[fin].len+1; 46 for(lsp=fin;lsp&&!s[lsp].tranc[c];lsp=s[lsp].pre) 47 s[lsp].tranc[c]=nwp; 48 if(!lsp) 49 s[nwp].pre=1; 50 else{ 51 lsq=s[lsp].tranc[c]; 52 if(s[lsq].len==s[lsp].len+1) 53 s[nwp].pre=lsq; 54 else{ 55 nwq=++siz; 56 s[nwq]=s[lsq]; 57 s[nwq].len=s[lsp].len+1; 58 s[nwp].pre=s[lsq].pre=nwq; 59 while(s[lsp].tranc[c]==lsq) 60 { 61 s[lsp].tranc[c]=nwq; 62 lsp=s[lsp].pre; 63 } 64 } 65 } 66 fin=nwp; 67 return ; 68 } 69 void Dfs(int x) 70 { 71 int t=p[x].blg; 72 for(int i=p[x].hd;i;i=e[i].lst) 73 { 74 int to=e[i].twd; 75 Dfs(to); 76 if(!t) 77 t=p[to].blg; 78 else if(t!=p[to].blg) 79 t=-1; 80 p[x].wgt+=p[to].wgt; 81 } 82 p[x].blg=t; 83 if(t<=0) 84 return ; 85 ans[t]+=(lnt)s[x].len-s[s[x].pre].len; 86 return ; 87 } 88 int main() 89 { 90 res(); 91 scanf("%d",&T); 92 for(int t=1;t<=T;t++) 93 { 94 fin=1; 95 lnt ans=0; 96 scanf("%s",tmp+1); 97 n=strlen(tmp+1); 98 for(int i=1;i<=n;i++) 99 Insert(tmp[i]-'a',t); 100 } 101 for(int i=2;i<=siz;i++) 102 ade(s[i].pre,i); 103 Dfs(1); 104 for(int i=1;i<=T;i++) 105 printf("%lld ",ans[i]); 106 return 0; 107 }