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\n", work(x, y));
}
}
return 0;
}