解题思路:看到题目，经典的区间查询+区间修改，我们用线段树维护一段区间的最小值，每当有新的订单，我们就先查询订单时间范围内的最小教室数量，然后与Dj作比较，如果比dj小，那么我们可以标记为false，然后后面的操作都不用看了，因为他只要求输出第一个不满足的订单，否则就按照订单进行区间修改,即在该订单的开始时间和结束时间这个范围内都减少dj个教室。

Code:

#include<bits/stdc++.h>
using namespace std;typedef long long ll;
const int INF = 0x3f3f3f3f3f3f3f3f;
const int N = 1000005;
ll a[N],tree[N<<2],lazy[N<<2];
int n,m;void pushup(int k) {tree[k] = min(tree[k<<1],tree[k<<1|1]);//将父节点更新为较小的值
}void pushdown(int k) {if(lazy[k]) {lazy[k<<1] += lazy[k];lazy[k<<1|1] += lazy[k];tree[k<<1] += lazy[k];tree[k<<1|1] += lazy[k];lazy[k] = 0;}
}void build(int l,int r,int k) {if(l == r) {tree[k] = a[l];}else {int mid = l + ((r - l) >> 1);build(l,mid,k<<1);build(mid+1,r,k<<1|1);pushup(k);}
}void updata(int L, int R, int v, int l, int r,int k) {if(L <= l && r <= R) {tree[k] += v;lazy[k] += v;}else {pushdown(k);int mid = l + ((r - l) >> 1);if(L <= mid) {updata(L,R,v,l,mid,k<<1);}if(R > mid) {updata(L,R,v,mid+1,r,k<<1|1);}pushup(k);}
}ll query(int L,int R,int l,int r,int k) {if(L <= l && r <= R) {return tree[k];}else {pushdown(k);ll res = INF;int mid = l + ((r - l) >> 1);if(L <= mid) {res = min(res,query(L,R,l,mid,k<<1));}if(R > mid) {res = min(res,query(L,R,mid+1,r,k<<1|1));}return res;}
}int main()
{int u,v;ll w;
//  freopen("X:\Âå¹È²âÊÔÊý¾Ý\P1083_2.in","r",stdin);scanf("%d%d",&n,&m);memset(lazy,0,sizeof lazy);for(int i = 1;i <= n; ++i) {scanf("%lld",&a[i]);}build(1,n,1);int fg = 0;//这个标记是否已经不满足条件for(int i = 1;i <= m; ++i) {scanf("%lld%d%d",&w,&u,&v);if(fg)continue;int k = query(u,v,1,n,1);if(k < w) {fg = i;}else {updata(u,v,-w,1,n,1);}}if(!fg) {puts("0");}else {printf("-1\n%d\n",fg);}return 0;
}