Description
Oimaster and sevenk love each other.
But recently,sevenk heard that a girl named ChuYuXun was dating with oimaster.As a woman's nature, s
evenk felt angry and began to check oimaster's online talk with ChuYuXun. Oimaster talked with Ch
uYuXun n times, and each online talk actually is a string.Sevenk asks q questions like this, "how
many strings in oimaster's online talk contain this string as their substrings?"
有n个大串和m个询问，每次给出一个字符串s询问在多少个大串中出现过
Input
There are two integers in the first line,
the number of strings n and the number of questions q.
And n lines follow, each of them is a string describing oimaster's online talk.
And q lines follow, each of them is a question.
n<=10000, q<=60000
the total length of n strings<=100000,
the total length of q question strings<=360000
Output
For each question, output the answer in one line.
Sample Input
3 3
abcabcabc
aaa
aafe
abc
a
ca
abcabcabc
aaa
aafe
abc
a
ca
Sample Output
1
3
3
1
解题思路：
利用Parent树的子节点都是父节点母串的性质，在大串的每个实节点打上标记，然后构建出这颗Parent树。
那么我们的问题就变成了在一个Fail节点子树内不同颜色个数。
像不像HH的项链？
把Dfs序构建出来，离线树状数组就好了。
代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 const int N=200000; 5 const int M=1000000; 6 struct sant{ 7 int tranc[26]; 8 int len; 9 int pre; 10 }s[N]; 11 struct pnt{ 12 int hd; 13 int col; 14 int ind; 15 int oud; 16 }p[N]; 17 struct ent{ 18 int twd; 19 int lst; 20 }e[N]; 21 struct qust{ 22 int l,r; 23 int ans; 24 int no; 25 }q[N]; 26 int siz; 27 int fin; 28 int n,Q; 29 int cnt; 30 int dfn; 31 char tmp[N]; 32 int line[N]; 33 int lst[N]; 34 int phr[N]; 35 int col[N]; 36 void Insert(int c,int whic) 37 { 38 int nwp,nwq,lsp,lsq; 39 nwp=++siz; 40 p[nwp].col=whic; 41 s[nwp].len=s[fin].len+1; 42 for(lsp=fin;lsp&&!s[lsp].tranc[c];lsp=s[lsp].pre) 43 s[lsp].tranc[c]=nwp; 44 if(!lsp) 45 s[nwp].pre=1; 46 else{ 47 lsq=s[lsp].tranc[c]; 48 if(s[lsq].len==s[lsp].len+1) 49 s[nwp].pre=lsq; 50 else{ 51 nwq=++siz; 52 s[nwq]=s[lsq]; 53 s[nwq].len=s[lsp].len+1; 54 s[nwp].pre=s[lsq].pre=nwq; 55 while(s[lsp].tranc[c]==lsq) 56 { 57 s[lsp].tranc[c]=nwq; 58 lsp=s[lsp].pre; 59 } 60 } 61 } 62 fin=nwp; 63 return ; 64 } 65 void ade(int f,int t) 66 { 67 cnt++; 68 e[cnt].twd=t; 69 e[cnt].lst=p[f].hd; 70 p[f].hd=cnt; 71 return ; 72 } 73 void Dfs(int x) 74 { 75 p[x].ind=++dfn; 76 phr[dfn]=x; 77 col[dfn]=p[x].col; 78 for(int i=p[x].hd;i;i=e[i].lst) 79 { 80 int to=e[i].twd; 81 Dfs(to); 82 } 83 p[x].oud=dfn; 84 return ; 85 } 86 bool cmp(qust x,qust y) 87 { 88 return x.r<y.r; 89 } 90 bool cmq(qust x,qust y) 91 { 92 return x.no<y.no; 93 } 94 int lowbit(int x) 95 { 96 return x&(-x); 97 } 98 void update(int pos,int x) 99 { 100 while(pos&&pos<=siz) 101 { 102 line[pos]+=x; 103 pos+=lowbit(pos); 104 } 105 return ; 106 } 107 int query(int pos) 108 { 109 int ans=0; 110 while(pos) 111 { 112 ans+=line[pos]; 113 pos-=lowbit(pos); 114 } 115 return ans; 116 } 117 int main() 118 { 119 fin=++siz; 120 scanf("%d%d",&n,&Q); 121 for(int i=1;i<=n;i++) 122 { 123 scanf("%s",tmp+1); 124 fin=1; 125 int len=strlen(tmp+1); 126 for(int j=1;j<=len;j++) 127 Insert(tmp[j]-'a',i); 128 } 129 for(int i=2;i<=siz;i++) 130 ade(s[i].pre,i); 131 Dfs(1); 132 133 for(int i=1;i<=Q;i++) 134 { 135 scanf("%s",tmp+1); 136 int root=1; 137 int len=strlen(tmp+1); 138 for(int j=1;j<=len;j++) 139 root=s[root].tranc[tmp[j]-'a']; 140 q[i].no=i; 141 if(!root) 142 continue; 143 q[i].l=p[root].ind; 144 q[i].r=p[root].oud; 145 } 146 std::sort(q+1,q+Q+1,cmp); 147 int rr=1; 148 for(int i=1;i<=Q;i++) 149 { 150 while(rr<=q[i].r) 151 { 152 if(!col[rr]) 153 { 154 rr++; 155 continue; 156 } 157 if(lst[col[rr]]) 158 update(lst[col[rr]],-1); 159 update(rr,1); 160 lst[col[rr]]=rr; 161 rr++; 162 } 163 rr--; 164 if(!q[i].l) 165 continue; 166 q[i].ans=query(q[i].r)-query(q[i].l-1); 167 } 168 std::sort(q+1,q+Q+1,cmq); 169 for(int i=1;i<=Q;i++) 170 printf("%d ",q[i].ans); 171 return 0; 172 }