(Analysis by Dhruv Rohatgi )

We essentially solve this problem with the two-pointers method and a bit of additional data structures.

Sort the boots and tiles by depth of snow, and maintain a doubly-linked list initially containing all the tiles. Iterate through the boots in order of decreasing snow resistance. For each boot, keep removing the tiles with the most snow from the doubly-linked list until the current pair of boots can withstand the depth of snow on all tiles in the linked list.

Then these boots are feasible for Farmer John if and only if the maximum distance between any adjacent pair of tiles in the linked list is no greater than the maximum distance which these boots can step. And as we remove tiles from the linked list, we can maintain this maximum distance: if it changes upon removing a tile, it must now be equal to the distance between the predecessor and successor of the deleted tile in the list.

This yields an $O(N\log N + B \log B)$ algorithm dominated by the cost of sorting.

#include <iostream>
#include <algorithm>
using namespace std;
#define MAXN 100000
#define MAXB 100000
int N,B;
int depth[MAXN];
int did[MAXN];
bool dcmp(int a,int b)
	return depth[a]<depth[b];
int snow[MAXB], dist[MAXB];
int ans[MAXB];
int bid[MAXB];
bool s_bcmp(int a,int b)
	return snow[a] < snow[b];
int nxt[MAXN];
int pre[MAXN];
int main()
	cin >> N >> B;
	for(int i=0;i<N;i++)
		cin >> depth[i];
		did[i] = i;
	for(int i=0;i<B;i++)
		cin >> snow[i] >> dist[i];
		bid[i] = i;
	for(int i=0;i<N;i++)
		nxt[i] = i+1, pre[i] = i-1;
	int j = N-1;
	int maxStep = 1;
	for(int i=B-1;i>=0;i--)
		int boot = bid[i];
		while(j >= 0 && depth[did[j]] > snow[boot])
			int cur = did[j];
			nxt[pre[cur]] = nxt[cur];
			pre[nxt[cur]] = pre[cur];
			maxStep = max(maxStep, nxt[cur] - pre[cur]);
		ans[boot] = (maxStep <= dist[boot]);
	for(int i=0;i<B;i++)
		cout << ans[i] << '\n';