【題解】LuoGu1083：借教室

原題傳送門
剛開始我想到了差分方法
可以$O(n)$檢驗是否所有訂單均可滿足
然後我就想到這道題是否和飛揚的小鳥一樣的套路再去用什麼方法找到第一個不能滿足的點

發現答案滿足二分性，那麼就直接二分好了，然後每次都用差分數組$O(n)\text{check}$
把所有均可滿足的情況一併包括進二分一起做

Code：

#include <bits/stdc++.h>
#define maxn 1000010
using namespace std;
long long delta[maxn];
int n, m, r[maxn], d[maxn], s[maxn], t[maxn];

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

bool check(int mid){
	for (int i = 1; i <= n; ++i) delta[i] = 0;
	for (int i = 1; i <= mid; ++i) delta[s[i]] += d[i], delta[t[i] + 1] -= d[i];
	for (int i = 1; i <= n; ++i)
		if ((delta[i] += delta[i - 1]) > r[i]) return 0;
	return 1;
}

int main(){
	n = read(), m = read();
	for (int i = 1; i <= n; ++i) r[i] = read();
	for (int i = 1; i <= m; ++i) d[i] = read(), s[i] = read(), t[i] = read();
	int l = 1, r = m, ans = -1;
	while (l <= r){
		int mid = (l + r) >> 1;
		if (check(mid)) l = mid + 1; else ans = mid, r = mid - 1;
	}
	if (ans == -1) puts("0"); else printf("-1\n%d\n", ans);
	return 0;
}