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<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
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;
}