Boring counting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3187 Accepted Submission(s): 1320

Problem Description

035 now faced a tough problem,his english teacher gives him a string,which consists with n lower case letter,he must figure out how many substrings appear at least twice,moreover,such apearances can not overlap each other.
Take aaaa as an example.”a” apears four times,”aa” apears two times without overlaping.however,aaa can’t apear more than one time without overlaping.since we can get “aaa” from [0-2](The position of string begins with 0) and [1-3]. But the interval [0-2] and [1-3] overlaps each other.So “aaa” can not take into account.Therefore,the answer is 2(“a”,and “aa”).

Input

The input data consist with several test cases.The input ends with a line “#”.each test case contain a string consists with lower letter,the length n won’t exceed 1000(n <= 1000).

Output

For each test case output an integer ans,which represent the answer for the test case.you’d better use int64 to avoid unnecessary trouble.

Sample Input

aaaa

ababcabb

aaaaaa

Sample Output

Source

2010 ACM-ICPC Multi-University Training Contest（9）——Host by HNU

思路：

　　枚举所有可能的长度，然后在height数组中求出这样的子串有多少个

　　复杂度O(n^2)

 1 #include <cstdlib>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 
 6 const int N = 200005;
 7 int sa[N],s[N],wa[N], wb[N], ws[N], wv[N];
 8 int rank[N], height[N];
 9 
10 bool cmp(int r[], int a, int b, int l)
11 {
12     return r[a] == r[b] && r[a+l] == r[b+l];
13 }
14 void calheight(int r[], int sa[], int n)
15 {
16     int i, j, k = 0;
17     for (i = 1; i <= n; ++i) rank[sa[i]] = i;
18     for (i = 0; i < n; height[rank[i++]] = k)
19         for (k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
20 }
21 void da(int r[], int sa[], int n, int m)
22 {
23     int i, j, p, *x = wa, *y = wb;
24     for (i = 0; i < m; ++i) ws[i] = 0;
25     for (i = 0; i < n; ++i) ws[x[i]=r[i]]++;
26     for (i = 1; i < m; ++i) ws[i] += ws[i-1];
27     for (i = n-1; i >= 0; --i) sa[--ws[x[i]]] = i;
28     for (j = 1, p = 1; p < n; j *= 2, m = p)
29     {
30         for (p = 0, i = n - j; i < n; ++i) y[p++] = i;
31         for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
32         for (i = 0; i < n; ++i) wv[i] = x[y[i]];
33         for (i = 0; i < m; ++i) ws[i] = 0;
34         for (i = 0; i < n; ++i) ws[wv[i]]++;
35         for (i = 1; i < m; ++i) ws[i] += ws[i-1];
36         for (i = n-1; i >= 0; --i) sa[--ws[wv[i]]] = y[i];
37         for (std::swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
38             x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
39     }
40     calheight(r,sa,n-1);
41 }
42 
43 int check(int lc,int x)
44 {
45     int mi=sa[1],mx=sa[1],ans=0;
46     for(int i=2;i<=lc;i++)
47     {
48         if(height[i]<x)
49         {
50             if(mx-mi>=x)
51                 ans++;
52             mx=mi=sa[i];
53         }
54         else
55         {
56             mx=std::max(mx,sa[i]);
57             mi=std::min(mi,sa[i]);
58         }
59     }
60     if(mx-mi>=x)ans++;
61     return ans;
62 }
63 char ss[N];
64 int main()
65 {
66     while(scanf("%s",ss)==1)
67     {
68         if(ss[0]=='#')break;
69         int la=strlen(ss),n=0,ans=0;
70         for(int i=0;i<la;i++)
71             s[n++]=ss[i]-'a'+1;
72         s[n]=0;
73         da(s,sa,n+1,30);
74         for(int i=1;i<=n/2;i++)
75            ans+=check(n,i);
76         printf("%d
",ans);
77     }
78     return 0;
79 }