A train runs on C Between two cities ( The departure city number is 1, The number of the city reached at the end is C), Suppose the train has S A seat , Now there is R Booking business . Now I want to know about this R This transaction is processed , See which reservations can meet , What can't be satisfied .

A reservation is made by O、D、N Three integers , From the starting station O To the target station D Need to book N A seat . A reservation can be satisfied means that there are empty seats within the travel range of the business , Otherwise, you can't meet . A business cannot be split , That is, the starting point and terminal cannot be changed , The number of reserved seats cannot be changed . All scheduled requirements are processed in the order given .

Please write a program , Look at those reservations that can meet , Those who can't meet .

        Input file The first line in the three integers C、S、R,(1<=c<=60 000, 1<=s<=60 000, 1<=r<=60 000) They are separated by a space . Next R Each line has three integers O、D、N,(1<=o<d<=c, 1<=n<=s), Represents each scheduled business separately .

        Right. I This business , If it can satisfy , In the output file in Of the I Line output “T”, Otherwise output “N”

4 6 4

1.

1 4 2

1.

1 3 2

1.

2 4 3

1.

1 2 3

T

1.

T

1.

N

1.

 N

This is a water problem with pits

The segment tree maintains the minimum value of the interval ,

You can't check and subtract , It is possible that the front meets the requirements , Less , The back doesn't meet the requirements

So if you judge to be satisfied , reduce , Otherwise, it will not be reduced

pit ：：： Order from l To r, The actual operation interval should be [l,r-1], Because I got off at the target station , Don't occupy a seat

#include<cstdio>
#include<algorithm>
#define N 60001
using namespace std;
int n,m,opl,opr,w,q;
bool ok;
class tree
{
private:
struct node
{
int l,r,minn,f;
}tr[N*4];
public:
void build(int k,int l,int r)
{
tr[k].l=l;tr[k].r=r;
if(l==r)
{
tr[k].minn=m;
return;
}
int mid=l+r>>1;
build(k<<1,l,mid);
build(k<<1|1,mid+1,r);
tr[k].minn=min(tr[k<<1].minn,tr[k<<1|1].minn);
}
void down(int k)
{
tr[k<<1].minn-=tr[k].f;
tr[k<<1|1].minn-=tr[k].f;
tr[k<<1].f+=tr[k].f;
tr[k<<1|1].f+=tr[k].f;
tr[k].f=0;
}
void solve(int k,int g)
{
if(tr[k].l>=opl&&tr[k].r<=opr)
{
if(g==1) { if(tr[k].minn<w) ok=true; }
else
{
tr[k].minn-=w;
tr[k].f+=w;
}
return;
}
if(tr[k].f) down(k);
int mid=tr[k].l+tr[k].r>>1;
if(opl<=mid) solve(k<<1,g);
if(opr>mid) solve(k<<1|1,g);
if(g==2) tr[k].minn=min(tr[k<<1].minn,tr[k<<1|1].minn);
}
}t;
int main()
{
scanf("%d%d%d",&n,&m,&q);
t.build(1,1,n);
for(int i=1;i<=q;i++)
{
ok=false;
scanf("%d%d%d",&opl,&opr,&w);
opr--;
t.solve(1,1);
if(ok) puts("N");
else
{
puts("T");
t.solve(1,2);
}
}
return 0;
}

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