SPOJ

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题意：

求一个字符串的不同子串的个数？

思路：

后缀数组算法：

每一个子串都是某一个后缀的前缀，那么问题转换为求所有后缀的不同前缀个数。

如果按照后缀数组 sa[0],sa[1],sa[2],,,sa[k]访问的话，

那么访问sa[k] 便会新加入n - sa[k] + 1个前缀。

这些前缀中有height[k] 是与前面重复的，减去即可。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1000 + 10;int n, k;int Rank[maxn];char s[maxn];int tmp[maxn ], sa[maxn], lcp[maxn];bool cmp(int i,int j){    if (Rank[i] != Rank[j]) return Rank[i] < Rank[j];    else {        int ri = i + k <= n ?  Rank[i+k] : -1;        int rj = j + k <= n ?  Rank[j + k]: -1;        return ri < rj;    }}void build_sa(char* s,int * sa){    for (int i = 0; i <= n; ++i){        sa[i] = i;        Rank[i] = i < n ? s[i] : -1;    }    for (k = 1; k <= n; k *= 2){        sort(sa, sa+n+1, cmp);        tmp[sa[0] ] = 0;        for (int i = 1; i <= n; ++i){            tmp[sa[i] ] = tmp[sa[i-1] ] + (cmp(sa[i-1], sa[i]) ? 1:0);        }        for (int i = 0; i <= n; ++i) {            Rank[i] = tmp[i];        }    }}void get_height(char* s, int* sa, int* lcp){    for (int i = 0; i <= n; ++i) Rank[sa[i] ] = i;    int h = 0;    lcp[0] = 0;    for (int i = 0; i < n; ++i){        int j = sa[Rank[i] -1 ];        if (h > 0) h--;        for (; j + h < n && i + h < n; ++h){            if (s[j+h] != s[i+h])break;        }        lcp[Rank[i]-1 ] = h;    }}int main(){    int T;    scanf("%d",&T);    while(T--){        memset(tmp,0,sizeof tmp);        memset(sa,0,sizeof sa);        memset(lcp,0,sizeof lcp);        memset(Rank,0,sizeof Rank);//        memset(tmp,0,sizeof tmp);        scanf("%s",s);        n = strlen(s);        build_sa(s, sa);        get_height(s,sa,lcp);        int ans = 0;//        for (int i = 1; i <= n; ++i) printf("%d %d\n", sa[i],lcp[i]);        for (int i = 1; i <= n; ++i){            ans += n-sa[i];            ans -= lcp[i];        }        printf("%d\n",ans);    }    return 0;}

DISUBSTR - Distinct Substrings

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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 <= 1000

Output

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

Example

Sample Input:
2
CCCCC
ABABA

Sample Output:
5
9

Explanation for the testcase with string ABABA:
len=1 : A,B
len=2 : AB,BA
len=3 : ABA,BAB
len=4 : ABAB,BABA
len=5 : ABABA
Thus, total number of distinct substrings is 9.

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