SPOJ Distinct Substrings（后缀数组求不同子串个数，好题）

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.

Submit solution!

 

题目链接：SPOJ DISUBSTR

一开始想用字典树，结果静态建树的Trie超时了（懒的写动态指针版……）真相是用后缀数组做的，因为每一个后缀的贡献原本为其长度，原本总贡献为$(len + 1) * len / 2$，但由于一些串重复，我们要减掉，再想一想，这些重复的是后缀的前缀，也就是$Suffix(x)$和$Suffix(y)$的公共前缀$LCP(x,y)$，但是x与y如何确定才能准确不遗漏地算出这些重复的串呢？按字典序排，然后height数组就是基于字典序排序的后缀，因此把所有height值减掉就好了。不过似乎有人用指针写的Trie过了，果然指针除了爆内存的风险，速度确实快啊。

想了一下用后缀数组只要$O(Nlog_{2}N)$，而字典树至少$O(N*N)$，果然不是一个档次……

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 1010;
int wa[N], wb[N], cnt[N], sa[N];
int ran[N], height[N];
char s[N];

inline int cmp(int r[], int a, int b, int d)
{
    return r[a] == r[b] && r[a + d] == r[b + d];
}
void DA(int n, int m)
{
    int i;
    int *x = wa, *y = wb;
    for (i = 0; i < m; ++i)
        cnt[i] = 0;
    for (i = 0; i < n; ++i)
        ++cnt[x[i] = s[i]];
    for (i = 1; i < m; ++i)
        cnt[i] += cnt[i - 1];
    for (i = n - 1; i >= 0; --i)
        sa[--cnt[x[i]]] = i;
    for (int k = 1; k <= n; k <<= 1)
    {
        int p = 0;
        for (i = n - k; i < n; ++i)
            y[p++] = i;
        for (i = 0; i < n; ++i)
            if (sa[i] >= k)
                y[p++] = sa[i] - k;
        for (i = 0; i < m; ++i)
            cnt[i] = 0;
        for (i = 0; i < n; ++i)
            ++cnt[x[y[i]]];
        for (i = 1; i < m; ++i)
            cnt[i] += cnt[i - 1];
        for (i = n - 1; i >= 0; --i)
            sa[--cnt[x[y[i]]]] = y[i];
        swap(x, y);
        x[sa[0]] = 0;
        p = 1;
        for (i = 1; i < n; ++i)
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], k) ? p - 1 : p++;
        m = p;
        if (p >= n)
            break;
    }
}
void getght(int n)
{
    int i, k = 0;
    for (i = 1; i <= n; ++i)
        ran[sa[i]] = i;
    for (i = 0; i < n; ++i)
    {
        if (k)
            --k;
        int j = sa[ran[i] - 1];
        while (s[i + k] == s[j + k])
            ++k;
        height[ran[i]] = k;
    }
}
int main(void)
{
    int T, i;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%s", s);
        int len = strlen(s);
        DA(len + 1, *max_element(s, s + len) + 1);
        getght(len);
        int ans = (len + 1) * len >> 1;
        for (i = 1; i <= len; ++i)
            ans -= height[i];
        printf("%d\n", ans);
    }
    return 0;
}