SPOJ 705 New Distinct Substrings

New Distinct Substrings

Time Limit: 2000ms

Memory Limit: 262144KB

This problem will be judged on  SPOJ. Original ID: SUBST1
64-bit integer IO format: %lld      Java class name: Main

 

Given a string, we need to find the total number of its distinct substrings.

Input

T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000

Output

For each test case output one number saying the number of distinct substrings.

Example

Input:

2

CCCCC

ABABA



Output:

5

9

Source

Base on a problem in ByteCode06

 

解题：求不同子串的个数。后缀数组lcp的应用。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 50010;

18 int rk[maxn],wb[maxn],wv[maxn],wd[maxn],lcp[maxn];

19 bool cmp(int *r,int i,int j,int k){

20     return r[i] == r[j] && r[i+k] == r[j+k];

21 }

22 void da(int *r,int *sa,int n,int m){

23     int i,k,p,*x = rk,*y = wb;

24     for(i = 0; i < m; ++i) wd[i] = 0;

25     for(i = 0; i < n; ++i) wd[x[i] = r[i]]++;

26     for(i = 1; i < m; ++i) wd[i] += wd[i-1];

27     for(i = n-1; i >= 0; --i) sa[--wd[x[i]]] = i;

28 

29     for(p = k = 1; p < n; k <<= 1,m = p){

30         for(p = 0,i = n-k; i < n; ++i) y[p++] = i;

31         for(i = 0; i < n; ++i) if(sa[i] >= k) y[p++] = sa[i] - k;

32         for(i = 0; i < n; ++i) wv[i] = x[y[i]];

33 

34         for(i = 0; i < m; ++i) wd[i] = 0;

35         for(i = 0; i < n; ++i) wd[wv[i]]++;

36         for(i = 1; i < m; ++i) wd[i] += wd[i-1];

37         for(i = n-1; i >= 0; --i) sa[--wd[wv[i]]] = y[i];

38 

39         swap(x,y);

40         x[sa[0]] = 0;

41         for(p = i = 1; i < n; ++i)

42             x[sa[i]] = cmp(y,sa[i-1],sa[i],k)?p-1:p++;

43     }

44 }

45 void calcp(int *r,int *sa,int n){

46     for(int i = 1; i <= n; ++i) rk[sa[i]] = i;

47     int h = 0;

48     for(int i = 0; i < n; ++i){

49         if(h > 0) h--;

50         for(int j = sa[rk[i]-1]; i+h < n && j+h < n; h++)

51             if(r[i+h] != r[j+h]) break;

52         lcp[rk[i]-1] = h;

53     }

54 }

55 int r[maxn],sa[maxn];

56 char str[maxn];

57 int main() {

58     int n;

59     scanf("%d",&n);

60     while(n--){

61         scanf("%s",str);

62         int len = strlen(str);

63         for(int i = 0; i < len; ++i)

64             r[i] = str[i];

65         r[len] = 0;

66         da(r,sa,len+1,128);

67         calcp(r,sa,len);

68         int ans = 0;

69         for(int i = 1; i <= len; ++i)

70             ans += len - sa[i] - lcp[i];

71         printf("%d\n",ans);

72     }

73     return 0;

74 }

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