Description
将一个字符串分成不超过kk个子串,在这些子串中选择一个字典序最大的子串(即子串的子串),称为“魔力串”,求一种分法使得“魔力串”字典序最小。
Solution
非常好的一道后缀数组的题!
最大的最小,显然要二分。
考虑二分魔力串的字典序排名,我们先通过排名找到原子串的位置,至于怎么判断一个子串是否合法,我们可以从后往前检查,如果加上当前后缀后字典序大于二分出来的子串,那么就在当前位置的后面切开,统计切开次数即可。
Source
/************************************************
* Au: Hany01
* Date: Apr 11th, 2018
* Prob: [BZOJ4310] 跳蚤
* Email: hany01@foxmail.com
************************************************/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia
template <typename T> inline bool checkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool checkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }
inline int read()
{
register int _, __; register char c_;
for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}
const int maxn = 100005;
int n, m, tp[maxn << 1], rk[maxn << 1], sa[maxn], c[maxn], f[maxn][18], Log[maxn];
char s[maxn];
inline void RadixSort()
{
For(i, 0, m) c[i] = 0;
For(i, 1, n) ++ c[rk[i]];
For(i, 1, m) c[i] += c[i - 1];
Fordown(i, n, 1) sa[c[rk[tp[i]]] --] = tp[i];
}
inline void getSA()
{
For(i, 1, n) rk[i] = s[i], tp[i] = i;
m = 255, RadixSort();
for (int k = 1, p; k <= n; k <<= 1)
{
p = 0;
For(i, n - k + 1, n) tp[++ p] = i;
For(i, 1, n) if (sa[i] > k) tp[++ p] = sa[i] - k;
RadixSort(), swap(rk, tp), rk[sa[1]] = 1, m = 1;
For(i, 2, n) rk[sa[i]] = tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + k] == tp[sa[i - 1] + k] ? m : ++ m;
if (m >= n) return;
}
}
inline void getHeight()
{
for (int i = 1, j, k = 0; i <= n; f[rk[i ++]][0] = k)
for (k = k ? k - 1 : 0, j = sa[rk[i] - 1]; s[i + k] == s[j + k] && i + k <= n && j + k <= n; ++ k) ;
For(i, 2, n) Log[i] = Log[i >> 1] + 1;
for (register int j = 1; j <= Log[n]; ++ j)
For(i, 1, n - (1 << j) + 1) f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
}
inline LL getSum()
{
LL Ans = 0;
For(i, 1, n) Ans += n - sa[i] - f[i][0] + 1;
return Ans;
}
inline PII kth(LL k)
{
For(i, 1, n)
if (k > n - f[i][0] - sa[i] + 1) k -= n - f[i][0] - sa[i] + 1;
else return mp(sa[i], sa[i] + f[i][0] + k - 1);
}
inline int LCP(int x, int y)
{
x = rk[x], y = rk[y];
if (x > y) swap(x, y);
if (x == y) return n - sa[y] + 1;
++ x;
register int t = Log[y - x + 1];
return min(f[x][t], f[y - (1 << t) + 1][t]);
}
inline bool compare(PII a, PII b)
{
//Return true if b > a
if (LCP(a.x, b.x) < min(a.y - a.x + 1, b.y - b.x + 1)) return rk[b.x] > rk[a.x];
else return b.y - b.x + 1 > a.y - a.x + 1;
}
inline int check(PII pos)
{
int cnt = 0, j;
for (int i = n; i >= 1; i = j)
{
for (j = i; j >= 1; -- j) if (compare(pos, mp(j, i))) break;
if (i == j) return INF;
++ cnt;
}
return cnt;
}
int main()
{
#ifdef hany01
File("bzoj4310");
#endif
static int k;
static LL L, R, M;
k = read(), scanf("%s", s + 1), n = strlen(s + 1);
getSA(), getHeight();
L = 1, R = getSum();
while (L < R)
if (check(kth(M = (L + R) >> 1)) <= k) R = M; else L = M + 1;
static PII pos = kth(L);
For(i, pos.x, pos.y) putchar(s[i]);
return 0;
}
//《秋夜曲》
//作者:王维
//桂魄初生秋露微,轻罗已薄未更衣。
//银筝夜久殷勤弄,心怯空房不忍归。