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P6095 [JSOI2015] 串分割

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P6095 [JSOI2015] 串分割

题面

题目链接

解法

显然分割的串长度不是$\lceil \frac{n}{k} \rceil$就是$\lfloor \frac{n}{k} \rfloor$。

不妨先破环为链，建立后缀数组并二分最大值后缀的排名$mx$，判断是否能将$n$个数划分成$k$段使得每段都$\le mx$。

首先枚举起始位置，设$len=\lceil \frac{n}{k} \rceil$，则一共最多有$len$个起始位置，然后每次贪心地划分，假设当前从$cur$位置开始划分，如果$s[cur…cur+len-1]\le mx$则划分$len$个，否则划分$len-1$个。重复$k$次看划分的总长度是否$\ge n$。判断的时间复杂度为$O(len \times \frac{n}{len})=O(n)$。

思考这种贪心的正确性。如果前一次划分$len$个，后一次由于$\gt mx$导致只划分了$len-1$个，总共就划分了$2\times len-1$个，容易发现这样和先划分$len-1$个再划分$len$个是一样的，所以正确性显然。故此问题能够在$O(n\log n)$的时间复杂度内解决。

AC代码

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/**
 * @file:           P6095.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P6095
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
    freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
    fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp& x) {
    bool sign = false;
    char ch = getchar();
    long double tmp = 1;
    for (; !isdigit(ch); ch = getchar())
        sign |= (ch == '-');
    for (x = 0; isdigit(ch); ch = getchar())
        x = x * 10 + (ch ^ 48);
    if (ch == '.')
        for (ch = getchar(); isdigit(ch); ch = getchar())
            tmp /= 10.0, x += tmp * (ch ^ 48);
    return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        write(x / 10);
    putchar((x % 10) ^ 48);
}
const int MAXN = 6e5 + 10;
const int LOGN = 19;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct SuffixArray {
    int n, sa[MAXN], rk[MAXN], tp[MAXN], he[MAXN];
    void radix_sort(int m) {
        static int buc[MAXN];
        for (int i = 0; i <= m; ++i)
            buc[i] = 0;
        for (int i = 1; i <= n; ++i)
            buc[rk[i]]++;
        for (int i = 1; i <= m; ++i)
            buc[i] += buc[i - 1];
        for (int i = n; i >= 1; --i)
            sa[buc[rk[tp[i]]]--] = tp[i];
    }
    void init(int n, char* s) {
        this->n = n;
        int m = 200;
        for (int i = 1; i <= n; ++i)
            rk[i] = s[i] + 1, tp[i] = i;
        radix_sort(m);
        for (int w = 1, p = 0; p < n; m = p, w <<= 1) {
            p = 0;
            for (int i = 1; i <= w; ++i)
                tp[++p] = n - w + i;
            for (int i = 1; i <= n; ++i)
                if (sa[i] > w)
                    tp[++p] = sa[i] - w;
            radix_sort(m);
            copy(rk + 1, rk + n + 1, tp + 1);
            rk[sa[1]] = p = 1;
            for (int i = 2; i <= n; ++i) {
                if (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w])
                    rk[sa[i]] = p;
                else
                    rk[sa[i]] = ++p;
            }
        }
        for (int i = 1, k = 0; i <= n; ++i) {
            if (k)
                k--;
            while (s[i + k] == s[sa[rk[i] - 1] + k])
                k++;
            he[rk[i]] = k;
        }
    }
} sa;
int n, k, ln;
int lg2[MAXN], st[MAXN][LOGN];
char str[MAXN];
int getlcp(int l, int r) {
    if(l > r)
        swap(l, r);
    if (l == r)
        return INF;
    int k = lg2[r - l];
    return min(st[l + 1][k], st[r - (1 << k) + 1][k]);
}
bool comp_lesseq(int x, int y, int len) {
    int lcp = getlcp(x, y);
    if (lcp >= len)
        return true;
    return str[sa.sa[x] + lcp] <= str[sa.sa[y] + lcp];
}
bool check(int mx) {
    for (int st = 1; st <= ln; ++st) {
        int cur = st;
        for (int i = 1; i <= k; ++i) {
            if (comp_lesseq(sa.rk[cur], mx, ln))
                cur += ln;
            else
                cur += ln - 1;
        }
        if (cur >= st + n)
            return true;
    }
    return false;
}
signed main() {
    read(n), read(k);
    scanf("%s", str + 1);
    copy(str + 1, str + n + 1, str + n + 1);
    sa.init(n + n, str);
    for (int i = 2; i <= n + n; ++i)
        lg2[i] = (i & (i - 1)) ? lg2[i - 1] : lg2[i - 1] + 1;
    for (int i = 1; i <= n + n; ++i)
        st[i][0] = sa.he[i];
    for (int j = 1; j < LOGN; ++j)
        for (int i = 1; i + (1 << j) - 1 <= n + n; ++i)
            st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
    ln = (n - 1) / k + 1;
    int l = 1, r = n + n, ans = r;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid))
            r = mid - 1, ans = mid;
        else
            l = mid + 1;
    }
    for (int i = 0; i < ln; ++i)
        putchar(str[sa.sa[ans] + i]);
    putchar('\n');
    return 0;
}