Distinct Substrings 【SPOJ - DISUBSTR】【後綴數組求不同子串個數】

題目鏈接

---

  求不同子串的個數。

  於是，我們可以考慮成，一個後綴，和它前一個sa的後綴有多少個不重疊的前綴子串。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e3 + 7, maxM = 3e3 + 10;
struct SA
{
    int n, m;
    int s[maxN];
    int y[maxN], x[maxN], c[maxM], sa[maxN], rk[maxN], height[maxN];
    inline void get_SA()
    {
        for(int i=1; i<=m; i++) c[i] = 0;   //桶的初始化
        for(int i=1; i<=n; i++) ++c[x[i] = s[i]];
        for(int i=2; i<=m; i++) c[i] += c[i - 1];   //利用差分前綴和的思想知道每個關鍵字最多是在第幾名
        for(int i=n; i>=1; i--) sa[c[x[i]]--] = i;
        for(int k=1; k<=n; k<<=1)
        {
            int num = 0;
            for(int i=n - k + 1; i<=n; i++) y[++num] = i;
            for(int i=1; i<=n; i++) if(sa[i] > k) y[++num] = sa[i] - k; //是否可以作爲第二關鍵字
            for(int i=1; i<=m; i++) c[i] = 0;
            for(int i=1; i<=n; i++) c[x[i]]++;  //因爲上一次循環已經求出這次的第一關鍵字了
            for(int i=2; i<=m; i++) c[i] += c[i - 1];
            for(int i=n; i>=1; i--) //在同一第一關鍵字下，按第二關鍵字來排
            {
                sa[c[x[y[i]]]--] = y[i];
                y[i] = 0;
            }
            swap(x, y);
            x[sa[1]] = 1; num = 1;
            for(int i=2; i<=n; i++)
            {
                x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) ? num : ++num;
            }
            if(num == n) break;
            m = num;
        }
    }
    inline void get_height()
    {
        int k = 0;
        for(int i=1; i<=n; i++) rk[sa[i]] = i;
        for(int i=1; i<=n; i++)
        {
            if(rk[i] == 1) continue;    //第一名的height爲0
            if(k) k--;  //height[i] >= height[i - 1] - 1
            int j = sa[rk[i] - 1];
            while(j + k <= n && i + k <= n && s[i + k] == s[j + k]) k++;
            height[rk[i]] = k;
        }
    }
    inline void clear()
    {
        n = 0; m = 307;
    }
} sa;
int N;
char s[maxN];
signed main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        sa.clear();
        scanf("%s", s + 1);
        N = (int)strlen(s + 1);
        for(int i=1; i<=N; i++) sa.s[++sa.n] = s[i];
        sa.s[++sa.n] = 300;
        sa.get_SA();
        sa.get_height();
        ll ans = 0;
        for(int i=1; i<=N; i++)
        {
            ans += (N - sa.sa[i] + 1) - sa.height[i];
        }
        printf("%lld\n", ans);
    }
    return 0;
}