CodeForces - 1296E2 String Coloring (hard version) dp

一、内容

This is a hard version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.You are given a string sconsisting of nlowercase Latin letters.You have to color all its characters the minimum number of colors (each character toexactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s). After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.The goal is to make the string sorted, i.e. all characters should be in alphabetical order.Your task is to find the minimum number of colors which you have to color the given string in so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.

Input

The first line of the input contains one integer n(1≤n≤2⋅105) — the length of s.The second line of the input contains the string sconsisting of exactly nlowercase Latin letters.

Output

In the first line print one integer res(1≤res≤n) — the minimum number of colors in which you have to color the given string so that after coloring it can become sorted by some sequence of swaps.In the second line print any possible coloring that can be used to sort the string using some sequence of swaps described in the problem statement. The coloring is the array cof length n, where 1≤ci≤res and ci means the color of the i-th character.

Input

9
abacbecfd

Output

2
1 1 2 1 2 1 2 1 2 

二、思路

-

三、代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 2e5 + 5;
int  n, dp[N], mxdp[26], rmax[26]; 
char s[N];
int main() {
	scanf("%d%s", &n, s + 1);
	for (int i = 1; i <= n; i++) {
		dp[i] = 1; //默认的最长长度是1 
		for (int j = 25; j > (s[i] - 'a'); j--) { //前面的最大下降子序列的长度
			dp[i] = max(dp[i], mxdp[j] + 1);
		} 
		mxdp[s[i] - 'a'] = max(dp[i], mxdp[s[i] - 'a']); //看是否能更新以i结尾的最长下降子序列 
	}
	int ans = 0;
	for (int i = 0; i < 26; i++) ans = max(ans, mxdp[i]);
	printf("%d\n", ans);
	for (int i = 1; i <= n; i++) printf("%d ", dp[i]);
	return 0;
}