SPOJ - SUBST1 D - New Distinct Substrings

时间：2025-04-03 22:04:25

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 ;

}

/*

*/

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