Description

很久很久以前,森林里住着一群跳蚤。一天,跳蚤国王得到了一个神秘的字符串,它想进行研究。首先,他会把串

分成不超过 k 个子串,然后对于每个子串 S,他会从S的所有子串中选择字典序最大的那一个,并在选出来的k个子串中选择字典序最大的那一个。他称其为“魔力串”。现在他想找一个最优的分法让“魔力串”字典序最小。

Input

第一行一个整数 k,K<=15

接下来一个长度不超过 10^5 的字符串 S。

Output

输出一行,表示字典序最小的“魔力串”。

Sample Input

2

ababa

Sample Output

ba

//解释:

分成aba和ba两个串,其中字典序最大的子串为ba


思路

首先我们要让所有段的最大子串的最大串最小,然后就可以考虑用二分,因为有一大堆子串的操作

不难想到后缀数组

然后就可以考虑怎么check

我们从后向前贪心

每次因为只需要向前扩展一个位置,所以每次只用检查一个子串是不是大于当前二分出的串

然后就可以很方便地做出来了


height处理的时候老是要写错

然后求lcp的时候注意把左区间的指针右移,并且要特判两个串的起始位置相等的情况

然后是贪心的时候每一段最后一个字符一定要特判


#include<bits/stdc++.h>

using namespace std;

typedef pair<int, int> pi;
typedef long long ll;
const int N = 1e5 + 10;
const int LOG = 20; struct Suffix_Array {
int s[N], n, m;
int c[N], x[N], y[N];
int height[N], sa[N], rank[N];
int st[N][LOG], Log[N];
ll rank_pre[N]; void init(int len, char *c) {
n = len, m = 0;
for (int i = 1; i <= len; i++) {
s[i] = c[i];
m = max(m, s[i]);
}
} void radix_sort() {
for (int i = 1; i <= m; i++) c[i] = 0;
for (int i = 1; i <= n; i++) c[x[y[i]]]++;
for (int i = 1; i <= m; i++) c[i] += c[i - 1];
for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];
} void buildsa() {
for (int i = 1; i <= n; i++) x[i] = s[i], y[i] = i;
radix_sort();
int now;
for (int k = 1; k <= n; k <<= 1) {
now = 0;
for (int i = n - k + 1; i <= n; i++) y[++now] = i;
for (int i = 1; i <= n; i++) if (sa[i] > k) y[++now] = sa[i] - k;
radix_sort();
y[sa[1]] = now = 1;
for (int i = 2; i <= n; i++) y[sa[i]] = (x[sa[i]] == x[sa[i - 1]] && x[sa[i] + k] == x[sa[i - 1] + k]) ? now : ++now;
swap(x, y);
if (now == n) break;
m = now;
}
} void buildrank() {
for (int i = 1; i <= n; i++) rank[sa[i]] = i;
} void buildrank_pre() {
for (int i = 1; i <= n; i++) rank_pre[i] = rank_pre[i - 1] + n - sa[i] + 1 - height[i];
} void buildheight() {
for (int i = 1; i <= n; i++) if (rank[i] != 1) {
int k = max(height[rank[i - 1]] - 1, 0); // 里面是 rank[i - 1]
for (; s[i + k] == s[sa[rank[i] - 1] + k]; k++);
height[rank[i]] = k; // height 里面是 rank
}
} void buildst() {
Log[1] = 0;
for (int i = 2; i < N; i++) Log[i] = Log[i >> 1] + 1;
for (int i = 1; i <= n; i++) st[i][0] = height[i];
for (int j = 1; j < LOG; j++) {
for (int i = 1; i + (1 << (j - 1)) <= n; i++) {
st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
}
}
} int queryst(int l, int r) {
if (l > r) swap(l, r);
++l; //***
int k = Log[r - l + 1];
return min(st[l][k], st[r - (1 << k) + 1][k]);
} int querylcp(int la, int ra, int lb, int rb) {
if (rank[la] == rank[lb]) return min(ra - la + 1, rb - lb + 1);
return min(min(ra - la + 1, rb - lb + 1), queryst(rank[la], rank[lb]));
} //return substringa <= substringb
bool cmpsubstring(int la, int ra, int lb, int rb) {
int lcp = querylcp(la, ra, lb, rb);
if (ra - la + 1 == lcp) return 1;
if (rb - lb + 1 == lcp) return 0;
return s[la + lcp] < s[lb + lcp];
} pi findkth(ll k) {
int pos = lower_bound(rank_pre + 1, rank_pre + n + 1, k) - rank_pre;
return pi(sa[pos], sa[pos] + height[pos] + k - rank_pre[pos - 1] - 1);
}
} Sa; int k, len;
char c[N]; bool check(pi cur) {
int last = len, tot = 0;
for (int i = len; i >= 1; i--) {
if (!Sa.cmpsubstring(i, last, cur.first, cur.second)) {
if (++tot > k) return 0;
last = i;
if (!Sa.cmpsubstring(i, last, cur.first, cur.second)) return 0;
}
}
return ++tot <= k;
} int main() {
#ifdef dream_maker
freopen("input.txt", "r", stdin);
#endif
scanf("%d", &k);
scanf("%s", c + 1);
len = strlen(c + 1);
Sa.init(len, c);
Sa.buildsa();
Sa.buildrank();
Sa.buildheight();
Sa.buildrank_pre();
Sa.buildst();
ll l = 1, r = Sa.rank_pre[len];
pi ans(1, 1);
while (l <= r) {
ll mid = (l + r) >> 1;
pi cur = Sa.findkth(mid);
if (check(cur)) {
ans = cur, r = mid - 1;
} else l = mid + 1;
}
for (int i = ans.first; i <= ans.second; i++) putchar(c[i]);
return 0;
}
05-11 20:07