P1083 [NOIP 2012 提高组] 借教室（差分）

说实话竟然没想到还能这样差分，这道题我们需要二分查找m个订单，对于每次二分用一次差分，然后看如果只考虑1到mid个订单是否会出现教室不够用的情况，如果够用说明导致教室不够用的订单在后面，应该让begin=mid+1；反之让end-1；其实这道题就是让我们找第一个出现教室不够用的订单

#include<iostream>
#include<set>
#include<map>
#include<algorithm>
#include<vector>
#include<cmath>
#include<climits>
#include<cstring>
#define int long long
const int N = 1e6+5;
using namespace std;
char* p1, * p2, buf[100000];
#define nc() (p1==p2 && (p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
int read()
{
	int x = 0, f = 1;
	char ch = nc();
	while (ch < 48 || ch>57)
	{
		if (ch == '-')
			f = -1;
		ch = nc();
	}
	while (ch >= 48 && ch <= 57)
		x = x * 10 + ch - 48, ch = nc();
	return x * f;
}
int a[N], b[N], c[N];
int p[N];
int cha[N];
int n, m;
int d[N], s[N], t[N];
int check() {
	int begin = 1, end = m;
	int mid;
	int res;
	while (begin <= end) {
		memset(b, 0, sizeof(b));
		mid = (begin + end) >> 1;
		for (int i = 1; i <= mid; i++) {
			b[s[i]] += d[i];
			b[t[i] + 1] -= d[i];
		}
		int p = 0;
		for (int i = 1; i <= n; i++) {
			b[i] += b[i - 1];
			if (b[i] > a[i]) {
				p = 1;
				break;
			}
		}
		if (p == 1)end=mid-1,res=mid;
		else begin=mid+1;
	}
	return res;
}
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for (int i = 1; i <= m; i++) {
		cin >> d[i] >> s[i] >> t[i];
		b[s[i]] += d[i];
		b[t[i] + 1] -= d[i];
	}
	int w = 0;
	for (int i = 1; i <= n; i++) {
		b[i] += b[i - 1];
		if (b[i] > a[i]) {
			w = 1; 
			break;
		}
	}
	if (w == 0)cout << 0;
	else cout << -1 << endl << check() << endl;
	
	return 0;
}