$dis[i][j]$ 表示节点 $j$到恐怖程度不超过 $i$的节点的最短距离.因 $i \le 100$,用$100$次BFS预处理即可.

#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
using namespace std;

const int MAXN = 1e5 + 5, MAXM = 105;
const int INF = 0x3f3f3f3f;

int n, m, q;  
int w[MAXN];
int dis[MAXM][MAXN]; 
vector<int> edges[MAXN];

void bfs(int a) { 
    queue<int> que;
    for (int i = 1; i <= n; i++) { 
        if (w[i] <= a) {
            que.push(i);
            dis[a][i] = 0;
        }
    }

    while (que.size()) {
        int u = que.front(); que.pop();
        for (auto v : edges[u]) {
            if (dis[a][v] > dis[a][u] + 1) {
                dis[a][v] = dis[a][u] + 1;
                que.push(v);
            }
        }
    }
}

void solve() {
    memset(dis, INF, sizeof(dis));

    cin >> n >> m >> q;
    for (int i = 1; i <= n; i++) cin >> w[i];
    while (m--) {
        int u, v; cin >> u >> v;
        edges[u].push_back(v), edges[v].push_back(u);
    }

    for (int i = 1; i <= 100; i++) bfs(i); 

    while (q--) {
        int p, a; cin >> p >> a;
        cout << (dis[a][p] == INF? -1 : dis[a][p]) << endl;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    solve();
    return 0;
}