[codeforces724D]Dense Subsequence

[codeforces724D]Dense Subsequence

试题描述

You are given a string s, consisting of lowercase English letters, and the integer m.

One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.

Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.

Formally, we choose a subsequence of indices 1 ≤ i_1 < i_2 < ... < i_t ≤ |s|. The selected sequence must meet the following condition: for every j such that 1 ≤ j ≤ |s| - m + 1, there must be at least one selected index that belongs to the segment [j,  j + m - 1], i.e. there should exist a k from 1 to t, such that j ≤ i_k ≤ j + m - 1.

Then we take any permutation p of the selected indices and form a new string s_{i_p1}s_{i_p2}... s_{i_pt}.

Find the lexicographically smallest string, that can be obtained using this procedure.

输入

The first line of the input contains a single integer m (1 ≤ m ≤ 100 000).

The second line contains the string s consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100 000. It is also guaranteed that the number m doesn't exceed the length of the string s.

输出

Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above.

输入示例

bcabcbaccba

输出示例

aaabb

数据规模及约定

见“输入”

题解

贪心贪心使劲贪。。。先暴力求一下必须要用到哪些字母，然后除掉最后一个字母，再用个堆和滑动窗口从左往右扫一遍，贪心地取刚才除掉的那个字母。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

int read() {
    int x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

#define maxn 100010
struct Node {
	char ch; int pos;
	Node() {}
	Node(char _, int __): ch(_), pos(__) {}
	bool operator < (const Node& t) const { return ch != t.ch ? ch > t.ch : pos < t.pos; }
	bool operator == (const Node& t) const { return ch == t.ch && pos == t.pos; }
} ;
priority_queue <Node> Q, delQ;
char S[maxn], ans[maxn];
int cnt, n, m;

bool vis[maxn];
void solve() {
	for(char ch = 'a'; ch <= 'z'; ch++) {
		for(int i = 1; i <= n; i++) if(S[i] == ch) vis[i] = 1, ans[++cnt] = ch;
		int lstp = 0; bool ok = 1;
		for(int i = 1; i <= n; i++) {
			if(vis[i]) lstp = i;
			if(i - lstp == m){ ok = 0; break; }
		}
		if(ok) {
			for(int i = 1; i <= n; i++) if(S[i] == ch) vis[i] = 0, cnt--;
			break;
		}
	}
	return ;
}

int main() {
	m = read();
	scanf("%s", S + 1); n = strlen(S + 1);

	solve();
	int lstp = 0;
	for(int i = 1; i < m; i++)
		if(vis[i]) lstp = i;
		else Q.push(Node(S[i], i));
	for(int i = m; i <= n; i++) {
		if(!vis[i]) Q.push(Node(S[i], i));
		else lstp = i;
//		printf("get: %d %d\n", i, lstp);
		if(i - lstp == m) {
			Node u = Q.top(); Q.pop();
			while(!delQ.empty() && u == delQ.top()) delQ.pop(), u = Q.top(), Q.pop();
			ans[++cnt] = u.ch; lstp = u.pos;
//			printf("set: %d\n", lstp);
		}
		if(i > m && !vis[i-m]) delQ.push(Node(S[i-m], i - m));
	}

	sort(ans + 1, ans + cnt + 1); ans[cnt+1] = 0;
	printf("%s\n", ans + 1);

	return 0;
}