Common Substrings
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 7605 | Accepted: 2524 |
Description
A substring of a string T is defined as:
T(i, k)=TiTi+1...Ti+k-1, 1≤i≤i+k-1≤|T|.
Given two strings A, B and one integer K, we define S, a set of triples (i, j, k):
S = {(i, j, k) | k≥K, A(i, k)=B(j, k)}.
You are to give the value of |S| for specific A, B and K.
Input
The input file contains several blocks of data. For each block, the first line contains one integer K, followed by two lines containing strings A and B, respectively. The input file is ended by K=0.
1 ≤ |A|, |B| ≤ 105
1 ≤ K ≤ min{|A|, |B|}
Characters of A and B are all Latin letters.
Output
For each case, output an integer |S|.
Sample Input
2 aababaa abaabaa 1 xx xx 0
Sample Output
22 5
题意:
给出两个串，问这两个串的所有的子串中（重复出现的，只要是位置不同就算两个子串），长度大于等于k的公共子串有多少个。
思路:
这个也是后缀自动机的比较好的题目.....
用一个串构建好后缀自动机,然后我们在这个串上面跑另外一个串,我们在这里应用上之前的spoj1811的匹配个数....
那么这样的串出现了多少次呢?....
就是spoj8222中的那个迭代更新的字串表示这个长度的子串出现了多少次的Right数组.....
我们这个东西就出现了 cnt * Right 次...真是太神奇了.....
1 #include<cstdlib>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #ifdef WIN32
6 #define fmt64 "%I64d"
7 #else
8 #define fmt64 "%lld"
9 #endif
10 using namespace std;
11 const int maxn = (int)2.5e5,sigma = 26;
12 char str[maxn];
13 int k;
14 int cmp(int, int);
15 struct Sam{
16 int ch[maxn][sigma << 1],par[maxn],stp[maxn],right[maxn],times[maxn],id[maxn];
17 int sz,last;
18 int idx(char x){
19 if(‘a‘ <= x && x <= ‘z‘) return x - ‘a‘;
20 else return x - ‘A‘ + 26;
21 }
22 void init(){
23 for(int i = 1; i <= sz; ++i){
24 memset(ch[i],0,sizeof(ch[i]));
25 right[i] = times[i] = par[i] = stp[i] = 0;
26 }
27 sz = last = 1;
28 }
29 void add(int c){
30 stp[++sz] = stp[last] + 1;
31 int p = last, np = sz;
32 for(; !ch[p][c]; p = par[p]) ch[p][c] = np;
33 if(!p) par[np] = 1;
34 else{
35 int q = ch[p][c];
36 if(stp[q] != stp[p] + 1){
37 stp[++sz] = stp[p] + 1;
38 int nq = sz;
39 memcpy(ch[nq],ch[q],sizeof(ch[q]));
40 par[nq] = par[q];
41 par[q] = par[np] = nq;
42 for(; ch[p][c] == q; p = par[p]) ch[p][c] = nq;
43 }
44 else par[np] = q;
45 }
46 last = np;
47 }
48 void ins(char *pt){
49 int i;
50 init();
51 for(i = 0; pt[i]; ++i) add(idx(pt[i]));
52 }
53 void prep(char *pt){
54 int i,x = 1;
55 for(i = 1; i <= sz; ++i) id[i] = i;
56 sort(id + 1, id + sz + 1, cmp);
57 for(i = 0; pt[i]; ++i){
58 x = ch[x][idx(pt[i])];
59 ++right[x];
60 }
61 for(i = 1; i <= sz; ++i)
62 if(par[id[i]]) right[par[id[i]]] += right[id[i]];
63 }
64 void comp(char *pt){
65 int i,x = 1,match = 0;
66 long long ans = 0;
67 for(i = 0; pt[i]; ++i){
68 if(ch[x][idx(pt[i])]){
69 x = ch[x][idx(pt[i])];
70 match++;
71 }
72 else{
73 while(x && !ch[x][idx(pt[i])]) x = par[x];
74 if(!x) x = 1, match = 0;
75 else match = stp[x] + 1, x = ch[x][idx(pt[i])];
76 }
77 if(match >= k){
78 ans += (long long)(match - max(k, stp[par[x]] + 1) + 1) * right[x];
79 if(par[x] && k <= stp[par[x]]) times[par[x]]++;
80 }
81 }
82 for(i = 1; i <= sz; ++i){
83 int t = id[i];
84 ans += (long long)times[t] * right[t] * (stp[t] - max(k, stp[par[t]] + 1) + 1);
85 if(par[t] && k <= stp[par[t]]) times[par[t]] += times[t];
86 }
87 printf(fmt64"\n",ans);
88 }
89 }sam;
90 int cmp(int x,int y){
91 return sam.stp[x] > sam.stp[y];
92 }
93 int main()
94 {
95 freopen("csb.in","r",stdin);
96 freopen("csb.out","w",stdout);
97 while(scanf("%d\n",&k) != EOF,k){
98 scanf("%s\n",str);
99 sam.ins(str);
100 sam.prep(str);
101 scanf("%s\n",str);
102 sam.comp(str);
103 }
104 return 0;
105 }