SPOJ 694 (后缀数组) Distinct Substrings

将所有后缀按照字典序排序后，每新加进来一个后缀，它将产生n - sa[i]个前缀。这里和小罗论文里边有点不太一样。

height[i]为和字典序前一个的LCP，所以还要减去，最终累计n - sa[i] - height[i]即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 100000 + 10;
char s[maxn];
int sa[maxn], rank[maxn], height[maxn];
int t[maxn], t2[maxn], c[maxn];
int n;

void build_sa(int n, int m)
{
    int i, *x = t, *y = t2;
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) c[x[i] = s[i]]++;
    for(i = 1; i < m; i++) c[i] += c[i - 1];
    for(i = n - 1; i >= 0; i--) sa[--c[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++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[y[i]]]++;
        for(i = 1; i < m; i++) c[i] += c[i - 1];
        for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for(i = 1; i < n; i++)
            x[sa[i]] = y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k] ? p - 1 : p++;
        if(p >= n) break;
        m = p;
    }
}

void build_height()
{
    int k = 0;
    for(int i = 1; i <= n; i++) rank[sa[i]] = i;
    for(int i = 0; i < n; i++)
    {
        if(k) k--;
        int j = sa[rank[i] - 1];
        while(s[i + k] == s[j + k]) k++;
        height[rank[i]] = k;
    }
}

int main()
{
    //freopen("in.txt", "r", stdin);

    int T; scanf("%d", &T);
    while(T--)
    {
        scanf("%s", s);
        n = strlen(s);
        memset(sa, 0, sizeof(sa));
        build_sa(n + 1, 256);
        build_height();

        int ans = 0;
        for(int i = 1; i <= n; i++)
            ans += n - sa[i] - height[i];
        printf("%d
", ans);
    }

    return 0;
}