不愧是IOI滴题，瞅了半天没思路/kk
贪心
将 (A)， (B) 机器人按照限制从小到大排序，物品按照质量从小到大排序，二分枚举最短时间，先用 (A) 机器人运，剩下的再交给 (B) 机器人，最后判断能不能运完，可以用优先队列来维护运的物品
/*
work by:Ariel_
队列要清空，无解输出-1
*/
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
const int N = 500005,M = 100005;
int read(){
int x = 0,f = 1; char c = getchar();
while(c < '0'||c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') {x = x*10 + c - '0'; c = getchar();}
return x*f;
}
struct node{
int x, y;
bool operator < (const node st) const {return x < st.x;}
}p[N];
int ta[M], tb[M], A, B, x[M], y[M], n;
priority_queue <int> Q;
bool check(int mid) {
for (int i = 1; i <= A; i++) ta[i] = mid;
for (int i = 1; i <= B; i++) tb[i] = mid;
while(!Q.empty()) Q.pop();
int now = 1;
for (int i = 1; i <= A; i++) {
while (p[now].x < x[i] && now <= n) Q.push(p[now].y), now++;//这个玩具可以被这个机器送（但不一定被这个机器送）
while (!Q.empty() && ta[i]--) Q.pop();//用这个机器运送这个玩具(要保证这个机器运送的时间在枚举范围之内)
}
while (now <= n) Q.push(p[now].y), ++now;
for (int i = B; i >= 1; i--) {
while (!Q.empty() && tb[i]-- && Q.top() < y[i]) Q.pop();
}
return Q.empty();
}
int main(){
A = read(), B = read(), n = read();
for (int i = 1; i <= A; i++) x[i] = read();
for (int i = 1; i <= B; i++) y[i] = read();
sort(x + 1, x + A + 1), sort(y + 1, y + B + 1);
for (int i = 1; i <= n; i++) p[i].x = read(), p[i].y = read(); sort(p + 1, p + n + 1);
int L = 1, R = n, ans = -1, mid;
while(L <= R) {
mid = (L + R) >> 1;
if (check(mid)) ans = mid, R = mid - 1;
else L = mid + 1;
}
printf("%d", ans);
puts("");
return 0;
}