LA 6856 Circle of digits 解题报告

题目链接

　　先用后缀数组给串排好序。dc3 O(n)

      二分答案+贪心check

　　答案的长度len=(n+k-1)/k

      如果起点为i长为len串大于当前枚举的答案,i的长度取len-1

　　从起点判断k个串的长度是否大于等于n

      check的时候最多枚举len个起点,每个位置需要枚举n/len个串,时间复杂度O(n)，总的时间复杂度O(nlogn+n)

#include <iostream>
#include <algorithm>
#include <string>
#include <fstream>
using namespace std;
#define ll long long
#define MAXN 1000009
int sa[MAXN], pos[MAXN], Rank[MAXN], r[MAXN];
int d, n, k;
string s;
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
int wa[MAXN], wb[MAXN], wv[MAXN], Ws[MAXN];
inline int c0 (int *r, int a, int b) {
    return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];
}
int c12 (int k, int *r, int a, int b) {
    if (k == 2) return r[a] < r[b] || r[a] == r[b] && c12 (1, r, a + 1, b + 1);
    else return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1];
}
inline void sort (int *r, int *a, int *b, int n, int m)
{
    int i;
    for (i = 0; i < n; i++) wv[i] = r[a[i]];
    for (i = 0; i < m; i++) Ws[i] = 0;
    for (i = 0; i < n; i++) Ws[wv[i]]++;
    for (i = 1; i < m; i++) Ws[i] += Ws[i - 1];
    for (i = n - 1; i >= 0; i--) b[--Ws[wv[i]]] = a[i];
    return;
}
inline void dc3 (int *r, int *sa, int n, int m)
{
    int i, j, *rn = r + n, *san = sa + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p;
    r[n] = r[n + 1] = 0;
    for (i = 0; i < n; i++) if (i % 3 != 0) wa[tbc++] = i;
    sort (r + 2, wa, wb, tbc, m);
    sort (r + 1, wb, wa, tbc, m);
    sort (r, wa, wb, tbc, m);
    for (p = 1, rn[F (wb[0])] = 0, i = 1; i < tbc; i++)
        rn[F (wb[i])] = c0 (r, wb[i - 1], wb[i]) ? p - 1 : p++;
    if (p < tbc) dc3 (rn, san, tbc, p);
    else for (i = 0; i < tbc; i++) san[rn[i]] = i;
    for (i = 0; i < tbc; i++) if (san[i] < tb) wb[ta++] = san[i] * 3;
    if (n % 3 == 1) wb[ta++] = n - 1;
    sort (r, wb, wa, ta, m);
    for (i = 0; i < tbc; i++) wv[wb[i] = G (san[i])] = i;
    for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
        sa[p] = c12 (wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
    for (; i < ta; p++) sa[p] = wa[i++];
    for (; j < tbc; p++) sa[p] = wb[j++];
    return;
}
inline bool check (int x) {
    for (int i = 0; i < d; i++) {
        int j = i, t = 0;
        while (t < k) {
            if (Rank[j % n] <= x) j += d;
            else
                j += d - 1;
            t++;
        }
        if (j - i >= n) return 1;
    }
    return 0;
}
int main() {
    while (scanf ("%d %d", &n, &k) == 2) {
        cin >> s; s += s;
        d = (n + k - 1) / k;
        for (int i = 0; i < (n << 1); i++) r[i] = s[i] - '0';
        r[n << 1] = 0;
        dc3 (r, sa, s.size() + 1, 10);
        for (int i = 1, p = 0; i <= (n << 1); i++)
            Rank[sa[i]] = ++p;
        for (int i = 1, p = 0; i <= (n << 1); i++)
            if (sa[i] < n ) pos[++p] = sa[i];
        int l = 1, r = n, last = -1;
        while (l <= r) {
            int mid = (l + r) >> 1;
            if (check (Rank[pos[mid]]) ) last = pos[mid], r = mid - 1;
            else
                l = mid + 1;
        }
        for (int i = last; i < last + d; i++)
            putchar (s[i]);
        putchar (10);
    }
}