这是一份代码：

#include <bits/stdc++.h>
using namespace std;
int n, m, k, cnt;
int dis[200005], p[200005], h[200005], id[200005];
vector <int> G[200005];
char buf[1 << 21], *p1 = buf, *p2 = buf;
#define getchar() p1 == p2 and (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++
inline int read() { char op = getchar(); int x = 0; bool flag = 0; while (op < 48 || op > 57) { if (op == '-') flag = 1; op = getchar(); } while (48 <= op && op <= 57) x = (x << 1) + (x << 3) + (op ^ 48), op = getchar(); return flag ? -x : x; }
inline void write(int x) { static int tmp[20], len = -1; if (x < 0) putchar('-'), x = -x; do tmp[++len] = x % 10, x /= 10; while (x); while(~len) putchar(tmp[len--] + 48); }
void bfs(int s, int h) {
	if (dis[s] >= h) return ;
	queue <pair <int, int>> q;
	q.push({s, h});
	dis[s] = h;
	while (!q.empty()) {
		pair <int, int> cur = q.front();
		q.pop();
		for (int i = 0; i < G[cur.first].size(); i++) {
			int v = G[cur.first][i];
			if (dis[v] >= cur.second - 1 or cur.second == 0) continue;
			dis[v] = cur.second - 1;
			q.push({v, cur.second - 1});
		}
	}
}
int main() {
	n = read(), m = read(), k = read();
	for (int i = 1; i <= n; i++) dis[i] = -1;
	for (int i = 1; i <= m; i++) {
		int u, v;
		u = read(), v = read();
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for (int i = 1; i <= k; i++) p[i] = read(), h[i] = read(), id[i] = i;
	sort(id + 1, id + 1 + k, [](int x, int y) { return h[x] > h[y]; });
	for (int i = 1; i <= k; i++) bfs(p[id[i]], h[id[i]]);
	for (int i = 1; i <= n; i++) {
		if (dis[i] != -1) cnt++;
	}
	write(cnt);
	putchar('\n');
	for (int i = 1; i <= n; i++) {
		if (dis[i] != -1) {
			write(i);
			putchar(' ');
		}
	}
    return 0;
}

由于超时了，加了一个特判，如果程序将要超时，就直接输出 $1$ 到 $n$。

#include <bits/stdc++.h>
using namespace std;
int n, m, k, cnt;
clock_t Start, End;
int dis[200005], p[200005], h[200005], id[200005];
vector <int> G[200005];
char buf[1 << 21], *p1 = buf, *p2 = buf;
#define getchar() p1 == p2 and (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++
inline int read() { char op = getchar(); int x = 0; bool flag = 0; while (op < 48 || op > 57) { if (op == '-') flag = 1; op = getchar(); } while (48 <= op && op <= 57) x = (x << 1) + (x << 3) + (op ^ 48), op = getchar(); return flag ? -x : x; }
inline void write(int x) { static int tmp[20], len = -1; if (x < 0) putchar('-'), x = -x; do tmp[++len] = x % 10, x /= 10; while (x); while(~len) putchar(tmp[len--] + 48); }
void bfs(int s, int h) {
	if (dis[s] >= h) return ;
	queue <pair <int, int>> q;
	q.push({s, h});
	dis[s] = h;
	while (!q.empty()) {
		pair <int, int> cur = q.front();
		End = clock();
		if (End - Start >= 500000) {
			write(n);
			putchar('\n');
			for (int i = 1; i <= n; i++) {
				write(i);
				putchar(' ');
			}
			exit(0);
		}
		q.pop();
		for (int i = 0; i < G[cur.first].size(); i++) {
			int v = G[cur.first][i];
			if (dis[v] >= cur.second - 1 or cur.second == 0) continue;
			dis[v] = cur.second - 1;
			q.push({v, cur.second - 1});
		}
	}
}
int main() {
	Start = clock();
	n = read(), m = read(), k = read();
	for (int i = 1; i <= n; i++) dis[i] = -1;
	for (int i = 1; i <= m; i++) {
		int u, v;
		u = read(), v = read();
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for (int i = 1; i <= k; i++) p[i] = read(), h[i] = read(), id[i] = i;
	sort(id + 1, id + 1 + k, [](int x, int y) { return h[x] > h[y]; });
	for (int i = 1; i <= k; i++) bfs(p[id[i]], h[id[i]]);
	for (int i = 1; i <= n; i++) {
		if (dis[i] != -1) cnt++;
	}
	write(cnt);
	putchar('\n');
	for (int i = 1; i <= n; i++) {
		if (dis[i] != -1) {
			write(i);
			putchar(' ');
		}
	}
    return 0;
}

于是，就这样AC了？