前缀和与差分。
由题意，每盏灯的照亮范围是区间 \([\max\{P-S,1\},\min\{P+S,N\}]\)。设 \(a_i\) 表示第 \(i\) 个停车位被几盏灯照亮，问题转化为对 \(L\) 个区间加一，最后查询 \(Q\) 个位置是否满足 \(a_i>0\)。
对于区间 \(l,r\)，将 \(\{a_N\}\) 的差分数组 \(\{d_N\}\) 对应位置 \(d_l\) 加一、\(d_{r+1}\) 减一，最后求前缀和，就可以得到数组 \(\{a_N\}\)，问题便迎刃而解了。
时间复杂度 \(O(N+L+Q)\)。
代码：
// Problem: P15540 [CCC 2026 J5/S2] Beams of Light
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P15540
// Memory Limit: 512 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(int x = (y); x <= (z); ++x)
#define per(x, y, z) for(int x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {uniform_int_distribution<int> dist(L, R);return dist(rnd);
}template<typename T> void chkmin(T& x, T y) {if(y < x) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}template<int mod>
inline unsigned int down(unsigned int x) {return x >= mod ? x - mod : x;
}template<int mod>
struct Modint {unsigned int x;Modint() = default;Modint(unsigned int x) : x(x) {}friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}friend Modint operator/(Modint a, Modint b) {return a * ~b;}friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}friend Modint operator~(Modint a) {return a ^ (mod - 2);}friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}friend Modint& operator++(Modint& a) {return a += 1;}friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}friend Modint& operator--(Modint& a) {return a -= 1;}friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}friend bool operator==(Modint a, Modint b) {return a.x == b.x;}friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};const int N = 5e5 + 5;int n, l, q, a[N];int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);cin >> n >> l >> q;rep(i, 1, l) {int p, s;cin >> p >> s;int l = max(p - s, 1), r = min(p + s, n);++a[l];--a[r + 1];}rep(i, 1, n) a[i] += a[i - 1];rep(t, 1, q) {int x;cin >> x;cout << (a[x] > 0 ? "Y" : "N") << endl;}return 0;
}