Luogu 1083（借教室）（NOIP 2012）

传送门

NOIP 2012 D2T2

题解：

二分答案（最早必须修改到的订单），差分维护前缀和，每次暴力check。复杂度O(nlog(n))，常数比线段树小，线段树加优化和卡常也能过。

#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int MAXN=1e6+4;
int n,m,x[MAXN],y[MAXN],ans=0;
ll a[MAXN],d[MAXN],sum[MAXN];
inline int read() {
	int x=0;char c=getchar();
	while (c<'0'||c>'9') c=getchar();
	while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x;
}
inline bool cck(int mid) {
	ll cur=0;
	memset(sum,0,sizeof(sum));
	for (register int i=1;i<=mid;++i) sum[x[i]]+=d[i],sum[y[i]+1]-=d[i];
	for (register int i=1;i<=n;++i) {
		cur+=sum[i];
		if (cur>a[i]) return false;
	}
	return true;
}
int main() {
//	freopen("P1083.in","r",stdin);
	n=read(),m=read();
	for (register int i=1;i<=n;++i) a[i]=read();
	for (register int i=1;i<=m;++i) d[i]=read(),x[i]=read(),y[i]=read();
	int l=1,r=n;
	while (l<=r) {
		int mid=(l+r)>>1;
		if (cck(mid)) l=mid+1;
		else ans=mid,r=mid-1;
	}
	if (!ans) puts("0");
	else printf("%d\n%d\n",-1,ans);
	return 0; 
}