NOI2016 区间

题目传送门

博客中为数不多的NOI的题目的题解……

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先将区间都离散化，再按区间长度排序
用双指针，并线段树维护出现最多的数出现了多少次
如果\(> m\)次，就移动左指针，删除最左边的区间，直到\(= m\)次，在这个过程中记录答案
如果\(< m\)次，就移动右指针，加入右边的区间
可以保证这样一定能找出最优解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
#define ls p << 1
#define rs p << 1 | 1
#define mid ((l+r) >> 1)
using namespace std;
LL read() {
	LL k = 0, f = 1; char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9')
		k = k * 10 + c - 48, c = getchar();
	return k * f;
}
struct zzz {
	int l, r, len;
}q[500010];
bool cmp(zzz x, zzz y) {
	return x.len < y.len;
}
int a[1000010], maxn = 0;
int tree[1000010 << 2], tag[1000010 << 2];
inline void up(int p) {
	tree[p] = max(tree[ls], tree[rs]);
}
inline void down(int p) {
	tree[ls] += tag[p]; tree[rs] += tag[p];
	tag[ls] += tag[p], tag[rs] += tag[p];
	tag[p] = 0;
}
void update(int nl, int nr, int k, int l = 1, int r = maxn, int p = 1) {
	if(nl <= l && nr >= r) {
		tree[p] += k; tag[p] += k; return ;
	}
	down(p);
	if(nl <= mid) update(nl, nr, k, l, mid, ls);
	if(nr > mid) update(nl, nr, k, mid+1, r, rs);
	up(p);
}
int main() {
	//freopen("2.in", "r", stdin);
	int n = read(), m = read();
	for(int i = 1; i <= n; ++i) {
		q[i].l = read(), q[i].r = read(), q[i].len = q[i].r - q[i].l;
		a[i << 1] = q[i].l, a[(i<<1) - 1] = q[i].r;
	}
	sort(a+1, a+n*2+1); int pos = unique(a+1, a+n*2+1) - a - 1;
	sort(q+1, q+n+1, cmp);
	for(int i = 1; i <= n; ++i)
		q[i].l = lower_bound(a+1, a+pos+1, q[i].l) - a,
		q[i].r = lower_bound(a+1, a+pos+1, q[i].r) - a,
		maxn = max(q[i].r, maxn);
	int l = 1, r = 0, ans = 2147483647;
	for(r = 1; r <= n; ++r) {
		update(q[r].l, q[r].r, 1);
		while(tree[1] >= m) {
			update(q[l].l, q[l].r, -1);
			ans = min(ans, q[r].len - q[l].len);
			++l;
		}
	}
	if(ans != 2147483647) cout << ans;
	else cout << -1;
	return 0;
}