CF1283E New Year Parties

\(\large{题目链接}\)
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\(\Large\textbf{Solution: } \large{考虑贪心。对于最少的，如果当前位置有人，那么把他往左移一定不会更差，所以就往左更优。\\对于最多的，首先在确保当前位置有人的情况下，优先往右更优，如果右边扩展了，那么再考虑向左扩展。}\)
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\(\Large\textbf{Code: }\)

#include <bits/stdc++.h>
using namespace std;

const int N = 2e5 + 5;

int n, a[N], b1[N], c[N], b2[N];

inline int read() {
	int x = 0;
	char ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return x; 
}

int main() {
	n = read();
	int x;
	for (int i = 1; i <= n; ++i) x = read(), ++a[x], ++c[x];
	int Min = 0, Max = 0, flg1 = 0, flg2 = 0;
	for (int i = 1; i <= n; ++i) {
		if (!flg1) {
			if (a[i] && !b1[i - 1]) ++b1[i - 1], --a[i];
			if (a[i]) ++b1[i], --a[i];
			if (a[i]) flg1 = 1, ++b1[i + 1];
		} else  {
			if (a[i]) ++b1[i + 1];
			else flg1 = 0;
		}
		if (!flg2) {
			if (c[i] && b2[i - 1]) c[i] = 0;
			if (c[i] && c[i + 1]) flg2 = 1, c[i] = 0, ++b2[i + 1];
			if (c[i]) ++b2[i + 1], c[i] = 0;
		} else c[i + 1] = 0, flg2 = 0;
	}
	for (int i = 1; i <= n + 1; ++i) 
		if (b1[i]) ++Max;
		if (b2[i]) ++Min;
	if (b1[0]) ++Max;
	printf("%d %d\n", Min, Max);
	return 0;
}