Codeforces 1165F2（二分内的check）

要点

• 二分答案，内部喜闻乐见的拖延策略：对于某个打折玩具，就选最晚的打折时间买，答案并不会变劣，只是购买时间的平移。
• 注意最晚时间不是预处理的东西，是二分内部的、在mid以内的最晚时间。

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn = 2e5 + 5;
int n, m, k[maxn], d[maxn], t[maxn];
int Late[maxn], sum;

bool ok(int mid) {
	for (int i = 1; i <= n; i++) {
		Late[i] = 0;
	}
	vector<int> day[maxn];
	for (int i = 1; i <= m; i++) {
		if (d[i] <= mid)
			Late[t[i]] = max(Late[t[i]], d[i]);
	}
	for (int i = 1; i <= n; i++) {
		if (k[i] && Late[i])
			day[Late[i]].push_back(i);
	}

	int s = sum, cur = 0;
	for (int i = 1; i <= mid; i++) {
		cur++;
		if (i < maxn && day[i].size()) {
			for (int t : day[i]) {
				if (cur >= k[t]) {
					cur -= k[t];
					s -= k[t];
				} else {
					s -= cur;
					cur = 0;
					break;
				}
			}
		}
	}
	return cur >= s * 2;
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)
		scanf("%d", &k[i]), sum += k[i];
	for (int i = 1; i <= m; i++) {
		scanf("%d%d", &d[i], &t[i]);
	}

	int l = 1, r = 4e5 + 5, ans;
	while (l <= r) {
		int mid = (l + r) >> 1;
		if (ok(mid))	ans = mid, r = mid - 1;
		else	l = mid + 1;
	}
	return !printf("%d\n", ans);
}