【VOJ1895】ニニスの守護后缀数组 DP

最新推荐文章于 2022-07-26 23:24:35 发布

hzt_Owen

最新推荐文章于 2022-07-26 23:24:35 发布

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分类专栏：字符串动态规划

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本文详细介绍了如何构造一个由递增数列组成的十进制数字符串，使得该字符串在字典序上尽可能大，并确保最后一个数是最小的。通过一系列算法操作实现这一目标，包括字符排序、哈希值计算和动态规划等技术。

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有一个十进制数字符串S，它是由一个严格上升的数列A拼接而成，要求你构造A使得：

1. 最后一个数最小

2. 在1的基础上字典序最大

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cassert>
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define Dwn(i, x, y) for (int i = x; i >= y; i --)
#define RepE(i, x) for (int i = pos[x]; i; i = g[i].nex)
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 100005;
const LL mod = 100000000000007LL, m2 = mod * 20;
int n, sa[N], t1[N], t2[N], rk[N], ht[N], c[N], f[N], g[N], ans, nx[N];
LL hs[N], sh[N];
char s1[N];
LL Mult(LL x, LL y) {
	LL z = 0;
	while (y) {
		if (y & 1) z += x;
		x <<= 1, y >>= 1;
		if (x > m2) x %= mod;
	}
	return z % mod;
}
bool cmp(int *r, int x, int y, int l) { return r[x] == r[y] && r[x + l] == r[y + l]; }
void Bsa(char *s, int n, int m) {
	int *x = t1, *y = t2, p = 0;
	Rep(i, 0, m) c[i] = 0;
	Rep(i, 0, n) c[ x[i] = s[i] ] ++;
	Rep(i, 1, m) c[i] += c[i - 1];
	Dwn(i, n, 0) sa[ --c[ x[i] ] ] = i;
	for (int j = 1; p <= n; j <<= 1, m = p) {
		p = 0;
		Rep(i, n-j+1, n) y[p ++] = i;
		Rep(i, 0, n) if (sa[i] >= j) y[p ++] = sa[i] - j;
		Rep(i, 0, m) c[i] = 0;
		Rep(i, 0, n) c[ x[y[i]] ] ++;
		Rep(i, 1, m) c[i] += c[i - 1];
		Dwn(i, n, 0) sa[ --c[ x[y[i]] ] ] = y[i];
		swap(x, y), p = 1, x[ sa[0] ] = 0;
		Rep(i, 1, n) x[ sa[i] ] = cmp(y, sa[i], sa[i - 1], j) ? p - 1 : p ++;
	}
}
void Bh(char *s, int n) {
	Rep(i, 0, n) rk[ sa[i] ] = i;
}
int main()
{
	scanf ("%s", s1), n = strlen(s1), s1[n] = 0;
	Bsa(s1, n, 300), Bh(s1, n), n --;
	f[0] = 0, sh[0] = 1;
	Rep(i, 1, n) sh[i] = sh[i - 1] * 37 % mod; // assert(sh[i] < 0);
	Dwn(i, n, 0) hs[i] = (hs[i + 1] * 37 + s1[i] - 'a') % mod; // 
	for (int i = 0, lt = 0; i <= n; i = lt) {
		while (s1[lt] == '0') lt ++;
		Rep(j, i, lt - 1) nx[j] = lt;
		if (s1[i] != '0') nx[i] = i, lt ++;
	}
	Rep(i, 0, n) {
		if (i) f[i] = max(f[i], f[i - 1]);
		 int k = f[i];
		f[i] = nx[ f[i] ];
		int l = i - f[i] + 1, j = nx[i + 1];
		LL h1, h2;
		h1 = hs[ f[i] ] - Mult(hs[ f[i] + l ], sh[l]) + mod;
		h2 = hs[j] - Mult(hs[j + l], sh[l]) + mod;
		if (rk[j] < rk[ f[i] ] || h1 % mod == h2 % mod) l ++;
		f[j + l - 1] = i + 1;
		 f[i] = k;
	}
	int p = f[n] - 1; g[ f[n] ] = n + 1;
	Dwn(i, n, 1) {
		if (s1[i] == '0') g[i] = max(g[i], g[i + 1]);
		if (!g[i]) continue ;
		int j = nx[i], l = g[i] - nx[i];
		if (i < l) { g[0] = max(g[0], i); break ; }
		LL h1, h2;
		h1 = hs[i - l] - Mult(hs[i], sh[l]) + mod;
		h2 = hs[j] - Mult(hs[j + l], sh[l]) + mod;
		if (rk[i - l] > rk[j] || h1 % mod == h2 % mod) l --;
		Rep(j, i - l, min(p, nx[ f[i - 1] ])) g[j] = i;
		p = min(p, i - l - 1);
	}
	while (ans <= n) {
		while (ans <= n && s1[ans] == '0') ans ++;
		if (!g[ans]) g[ans] = n + 1;
		Rep(i, ans, g[ ans ] - 1) printf("%c", s1[i]);
		printf(" "); ans = g[ans];
	}
	puts("");

	return 0;
}