题目链接

题意

对于给定的字符串，求有多少个 不重叠的子串 出现次数 \(\geq 2\).

思路

枚举子串长度 \(len\)，以此作为分界值来对 \(height\) 值进行划分。

显然，对于每一组，组内子串具有一个长度为 \(len\) 的公共前缀。

至于是否重叠，只需判断 \(sa_{max}-sa_{min}\geq len\).

对于组间，显然它们的公共前缀互不相同，所以答案即为\(\sum_{len}\sum_{group}\)

Code

#include <bits/stdc++.h>

#define maxn 1010

using namespace std;

typedef long long LL;

int wa[maxn], wb[maxn], wv[maxn], wt[maxn], h[maxn], rk[maxn], sa[maxn], n, r[maxn];

char s[maxn];

char s1[maxn], s2[maxn];

bool cmp(int* r, int a, int b, int l) { return r[a] == r[b] && r[a+l] == r[b+l]; }

void init(int* r, int* sa, int n, int m) {

    int* x=wa, *y=wb, *t, i, j, p;

    for (i = 0; i < m; ++i) wt[i] = 0;

    for (i = 0; i < n; ++i) ++wt[x[i] = r[i]];

    for (i = 1; i < m; ++i) wt[i] += wt[i - 1];

    for (i = n-1; i >= 0; --i) sa[--wt[x[i]]] = i;

    for (j = 1, p = 1; p < n; j <<= 1, m = p) {

        for (p = 0, i = n-j; i < n; ++i) y[p++] = i;

        for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;

        for (i = 0; i < n; ++i) wv[i] = x[y[i]];

        for (i = 0; i < m; ++i) wt[i] = 0;

        for (i = 0; i < n; ++i) ++wt[wv[i]];

        for (i = 1; i < m; ++i) wt[i] += wt[i - 1];

        for (i = n-1; i >= 0; --i) sa[--wt[wv[i]]] = y[i];

        t = x, x = y, y = t, x[sa[0]] = 0;

        for (p = 1, i = 1; i < n; ++i) x[sa[i]] = cmp(y, sa[i], sa[i-1], j) ? p - 1 : p++;

    }

    for (i = 0; i < n; ++i) rk[sa[i]] = i;

    int k = 0;

    for (i = 0; i < n - 1; h[rk[i++]] = k) {

        for (k = k ? --k : 0, j = sa[rk[i] - 1]; r[i+k] == r[j+k]; ++k);

    }

}

void work() {

    int tot=0, m=0;

    for (int i = 0; s[i]; ++i) m = max(r[tot++] = s[i], m);

    r[tot++] = 0;

    init(r, sa, tot, ++m);

    int ans = 0;

    for (int len = 1; len < (tot+1)/2; ++len) {

        int minn = maxn, maxx = 0;

        for (int i = 1; i < tot; ++i) {

            if (h[i] < len) {

                ans += maxx-minn >= len ? 1 : 0;

                minn = sa[i], maxx = sa[i];

            }

            else minn = min(minn, sa[i]), maxx = max(maxx, sa[i]);

        }

        ans += maxx-minn >= len ? 1 : 0;

    }

    printf("%d\n", ans);

}

int main() {

    while (scanf("%s", s) != EOF && s[0] != '#') work();

    return 0;

}

巴特西

hdu 3518 Boring counting 后缀数组 height分组

题目链接

题意

思路

Code

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