SPOJ - SUBST1 D - New Distinct Substrings

D - New Distinct Substrings

题目大意：求一个字符串中不同子串的个数。

裸的后缀数组

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N = 1e5 + ;
const int M = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;

char s[N];
int sa[N], t[N], t2[N], c[N], rk[N], height[N], id[N], b[N], d[N], n, tot;

void buildSa(char *s, int n, int m) {
    int i, j = , k = , *x = t, *y = t2;
    for(i = ; i < m; i++) c[i] = ;
    for(i = ; i < n; i++) c[x[i] = s[i]]++;
    for(i = ; i < m; i++) c[i] += c[i - ];
    for(i = n - ; i >= ; i--) sa[--c[x[i]]] = i;
    for(int k = ; k <= n; k <<= ) {
        int p = ;
        for(i = n - k; i < n; i++) y[p++] = i;
        for(i = ; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = ; i < m; i++) c[i] = ;
        for(i = ; i < n; i++) c[x[y[i]]]++;
        for(i = ; i < m; i++) c[i] += c[i - ];
        for(i = n - ; i >= ; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x, y);
        p = ; x[sa[]] = ;
        for(int i = ; i < n; i++) {
            if(y[sa[i - ]] == y[sa[i]] && y[sa[i - ] + k] == y[sa[i] + k])
                x[sa[i]] = p - ;
            else x[sa[i]] = p++;
        }
        if(p >= n) break;
        m = p;
     }

     for(i = ; i < n; i++) rk[sa[i]] = i;
     for(i = ; i < n - ; i++) {
        if(k) k--;
        j = sa[rk[i] - ];
        while(s[i + k] == s[j + k]) k++;
        height[rk[i]] = k;
     }
}

int main() {
    int T; scanf("%d", &T);
    while(scanf("%s", s) != EOF) {

        n = strlen(s);
        buildSa(s, n + , );

        LL ans = ;

        for(int i = ; i <= n; i++) {
            ans += (n - sa[i]) - height[i];
        }

        printf("%lld\n", ans);
    }
    return ;
}

/*
*/