Codeforces 1175E Minimal Segment Cover

题意：
有(n)条线段，区间为([l_i, r_i])，每次询问([x_i, y_i])，问要被覆盖最少要用多少条线段。

思路：
(f[i][j])表示以(i)为左端点，用了(2^j)条线段，最远到哪里。
然后从大到小贪心即可，类似于倍增找LCA的过程。

代码：

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

#define N 200010
#define M 500010
#define D 20
int n, q;
int l[N], r[N];
int f[M][D];

int work(int x, int y) {
	int ans = 0;
	for (int i = D - 1; i >= 0; --i) {
		if (f[x][i] < y) {
			x = f[x][i];
			ans |= (1 << i);
		}
	}
	x = f[x][0]; ++ans;
	if (x < y) ans = -1;
	return ans;
}

int main() {
	while (scanf("%d%d", &n, &q) != EOF) {
		for (int i = 1; i <= n; ++i) {
			scanf("%d%d", l + i, r + i);
			++l[i], ++r[i];
		}
		memset(f, 0, sizeof f);
		for (int i = 1; i <= n; ++i) {
			f[l[i]][0] = max(f[l[i]][0], r[i]);
		}
		for (int i = 1; i < M; ++i) {
			f[i][0] = max(f[i][0], max(i, f[i - 1][0]));
			for (int j = 1; j < D; ++j) {  
				f[i][j] = max(f[i][j], max(f[i][j - 1], f[i - 1][j]));
			}
		}
	
		for (int j = 1; j < D; ++j) {
			for (int i = 1; i < M; ++i) {
				f[i][j] = max(f[i][j], f[f[i][j - 1]][j - 1]);
			}
		}	
		int x, y;
		while (q--) {
			scanf("%d%d", &x, &y);
			++x, ++y;
			printf("%d
", work(x, y));
		}
	}
	return 0;
}