BZOJ 3048USACO2013 JanCow Lineup 滑块思想

Posted 2020-07-01

昨天下午想了好久没想出来，果然是很弱，思考能力低下。

用的类似单调队列的思想，维护一个长度为$k+1$的滑块，每次统计下$ans$就可以了

#include<cstdio>
#include<algorithm>
using namespace std;
int n, k, H[100003], id[100003], a[100003], cnt = 0, c[100003], st, ans = 0;
inline bool cmp(int X, int Y) {
	return H[X] < H[Y];
}
int main() {
	scanf("%d%d\n", &n, &k);
	for(int i = 1; i <= n; ++i) {
		scanf("%d\n", &H[i]);
		id[i] = i;
	}
	id[0] = 0;
	H[0] = 100003;
	sort(id + 1, id + n + 1, cmp);
	for(int i = 1; i <= n; ++i) {
		if (H[id[i]] != H[id[i-1]])
			++cnt;
		a[id[i]] = cnt;
	}
	cnt = 0;
	st = 0;
	for(int i = 1; i <= n; ++i) {
		++c[a[i]];
		if (c[a[i]] == 1) {
			++cnt;
			while (cnt > k + 1) {
				--c[a[st]];
				if (c[a[st]] == 0)
					--cnt;
				++st;
			}
		}
		ans = max(ans, c[a[i]]);
	}
	printf("%d\n",ans);
	return 0;
}

没了