思维题。
以地板为序构造链表，再排序，然后删除走不过去的地面。
删除的时候顺便维护最大的跨度，以此判断可行性。
总的来说利用了答案的单调性。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 1e5 + 20;
inline int read()
{int x = 0; char ch = getchar(); bool f = false;while(!isdigit(ch)) f |= (ch == '-'), ch = getchar();while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();return f ? -x : x;
}int N, B;
struct Node
{int val;Node *pre, *nxt;
}node[MAXN];struct boot
{int dep, dis, idx, ans;inline bool operator >(const boot &rhs) const {return dep > rhs.dep;}inline bool operator <(const boot &rhs) const {return idx < rhs.idx;}
}b[MAXN];
struct Floor
{int dep, idx;Node *pos;inline bool operator >(const Floor &rhs) const {return dep > rhs.dep;}inline bool operator <(const Floor &rhs) const {return idx < rhs.idx;}
}f[MAXN];void build(){f[1].pos = &node[1]; node[1].val = 1;for(int i = 2; i <= N; i++){node[i].pre = &node[i - 1], node[i - 1].nxt = &node[i];node[i].val = 1, f[i].pos = &node[i];}
}int main()
{cin>>N>>B;for(int i = 1; i <= N; i++) f[i] = (Floor){read(), i};for(int i = 1; i <= B; i++) b[i].dep = read(), b[i].dis = read(), b[i].idx = i;build();sort(f + 1, f + N + 1, greater<Floor>());sort(b + 1, b + B + 1, greater<boot>());int p = 1, maxs = 1;for(int i = 1; i <= B; i++){while(p <= N && f[p].dep > b[i].dep) {Node *cur = f[p].pos;cur->pre->nxt = cur->nxt;cur->nxt->pre = cur->pre;maxs = max(maxs, cur->pre->val += cur->val);++p;}if(maxs > b[i].dis) b[i].ans = 0;else b[i].ans = 1;}sort(b + 1, b + B + 1);for(int i = 1; i <= B; i++) printf("%d\n", b[i].ans);return 0;
}
