$Luogu$

 

$Sol$

区间整体加减? 差分+树状数组维护前缀和!

那每给一个人借完教室之后都要判断一下现在合不合法?那复杂度比暴力还不如些...

注意到这里的单调性,假设给前$x$个人借完教室之后就不合法了,那给前$x+1,x+2.....$个人借教室一定也是不合法的.于是可以二分第一个需要修改申请的人,然后$check()$.$check()$里在借完所有的人教室之后再判断合不合法就好了,而且就用不着树状数组了.

$over.$

 

$Code$

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define il inline
#define Rg register
#define go(i,a,b) for(Rg int i=a;i<=b;++i)
#define yes(i,a,b) for(Rg int i=a;i>=b;--i)
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define db double
#define inf 2147483647
using namespace std;
il int read()
{
    Rg int x=0,y=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*y;
}
const int N=1000010;
int n,m,r[N],d[N],s[N],t[N],a[N],as;
il bool ck(int x)
{
    mem(a,0);
    go(i,1,x)
    {
        a[s[i]]+=d[i];
        a[t[i]+1]-=d[i];
    }
    Rg int nw=0;
    go(i,1,n){nw+=a[i];if(nw>r[i])return 1;}
    return 0;
}
int main()
{
    n=read(),m=read();
    go(i,1,n)r[i]=read();
    go(i,1,m)d[i]=read(),s[i]=read(),t[i]=read();
    Rg int l=1,r=m,mid;
    while(l<=r)
    {
        mid=(l+r)>>1;//cout<<l<<" "<<r<<" "<<mid<<endl;
        if(ck(mid))as=mid,r=mid-1;
        else l=mid+1;
    }
    if(!as)printf("0\n");
    else printf("-1\n%d\n",as);
    return 0;
}

View Code