rt,在其他OJ和本地运行样例一都没问题，但提交就全TLE，不开O2就RE。

思路：用把深度大于s的块置1，旁边也有1就联通在一起，并查集来维护连通块最大长度，某个鞋子能通过必须大于它。

int a[]:雪地深度，PII c[]升序雪地深度和编号 b[] 深度s,步长d,编号id

#include<bits/stdc++.h>
#include<unordered_map>
typedef long long ll;
typedef long double ld;
#define debug(a) cout<<a<<'\n'
#define Pll pair<ll,ll>
#define PII pair<int,int>
#define ft first
#define sd second
#define vec vector
#define pushk push_back
#define pl p<<1
#define pr p<<1|1
using namespace std;
const int N = 1e5+ 10, M = 300, mod = 1e6;
const ll inf = 1e18;
const ld eps = 1e-8;
int mov[4][2] = { {0,1},{1,0},{-1,0},{0,-1} }, dx, dy;
void bout(bool f) {
    if (f)cout << "Yes\n";
    else cout << "No\n";
}
ll n, m, k;
int a[N],ans[N];
PII c[N];
struct node{
	int s,d,id;
	bool operator<(const node n)const {
		return s<n.s;
	}
};
node b[N];

int fa[N],rk[N];
int find(int x){
	if(x==fa[x])return x;
	else return fa[x]=find(fa[x]);
}

bool unite(int x,int y){
	x=find(x);
	y=find(y);
	if(x==y)return 0;
	rk[x]+=rk[y];
	fa[y]=x; 
}
bool st[N];

void solve(){
	cin>>n>>m;
	for(int i=1;i<=n;++i){
		fa[i]=i;
		rk[i]=1;
	}
	for(int i=1;i<=n;++i){
		cin>>a[i];
		c[i]={a[i],i};
	}
	sort(c+1,c+n+1);
	for(int i=1;i<=m;++i){
		cin>>b[i].s>>b[i].d;
		b[i].id=i;
	}
	sort(b+1,b+m+1);
	int p=n,res=0,x,id;
	for(int i=m;i>=1;--i){
		while(b[i].s<c[p].ft){
			id=c[p].sd;
			--p;
			st[id]=1;
			if(st[id-1])unite(id,id-1);
			if(st[id+1])unite(id,id+1);
			res=max(res,rk[find(id)]);
		}
		if(b[i].d>res)ans[b[i].id]=1;
		else ans[b[i].id]=0;
	}
	for(int i=1;i<=m;++i)cout<<ans[i]<<'\n';
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifndef ONLINE_JUDGE
    streambuf* cinbackup = cin.rdbuf(), * coutbackup = cout.rdbuf();
    ifstream fin("in.txt");
    ofstream fout("out.txt");
    cin.rdbuf(fin.rdbuf());
    cout.rdbuf(fout.rdbuf());
#endif
    int _ = 1, __ = 1;
    //cin >> _;
    __ = _;
    while (_--) {
        solve();
    }
    return 0;
}