線段樹水題

我們先建一棵線段樹,對於每次借教室相當於區間減,等哪天減到小於零就輸出就行了

這題據說nm暴力可過

程式碼

//By AcerMo
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int M=4000005;
#define ls t[x].s[0]
#define rs t[x].s[1]
#define lb t[x].b[0]
#define rb t[x].b[1]
 

struct Seg
{
    int s[2],b[2];
    int asum,lazy;
}t[M];
int n,m;
int a[M],cnt,rt;
bool flag=0;
inline int read()
{
    int x=0;char ch=getchar();
    while (ch>'9'||ch<'0') ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x;
}
inline void up(int x)
{
    t[x].asum=min(t[ls].asum+t[x].lazy,t[rs].asum+t[x].lazy);
    if
 
 (t[x].asum<0) flag=1;
    return ;
}
inline void built(int l,int r,int x)
{
    lb=l;rb=r;t[x].lazy=0;
    if (l==r)
    {
        ls=rs=-1;t[x].asum=a[l];
        return ;
    }
    int mid=(l+r)>>1;
    ls=++cnt;rs=++cnt;
    built(l,mid,ls);built(mid+1,r,rs);
    return (void)(up(x));
}
inline
 
 void date(int l,int r,int x,int v)
{
    if (l<=lb&&rb<=r) 
    {
        t[x].asum+=v;t[x].lazy+=v;
        if (t[x].asum<0) flag=1;
        return ;
    }
    int mid=(lb+rb)>>1;
    if (l<=mid) date(l,r,ls,v);
    if (r>mid) date(l,r,rs,v);  
    return (void)(up(x));
}
signed main()
{
    n=read();m=read();
    for (int i=1;i<=n;i++) a[i]=read();
    built(1,n,rt);
    for (int i=1;i<=m;i++)
    {
        int sub=read(),l=read(),r=read();
        date(l,r,rt,-sub);
        if (flag) return cout<<"-1"<<endl<<i,0; 
    }
    return puts("0"),0;
}