CF1555E Boring Segments(尺取+线段树)

// Problem: E. Boring Segments
// Contest: Codeforces - Educational Codeforces Round 112 (Rated for Div. 2)
// URL: http://codeforces.com/contest/1555/problem/E
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
  
inline void out(ll x){
    if (x < 0) x = ~x + 1, putchar('-');
    if (x > 9) out(x / 10);
    putchar(x % 10 + '0');
}
  
inline void write(ll x){
    if (x < 0) x = ~x + 1, putchar('-');
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
  
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)res = res * a ;
        a = a * a ;
        b >>= 1;
    }
    return res;
}

const int inf = 0x3f3f3f3f;
const int maxn=1e6+7,maxm=210000;

int n,m;
struct line{
	int l,r;
	ll w;
}p[maxn];

bool cmp(line a,line b){
	return a.w<b.w;
}

struct node{
	int l,r;
	ll laz,minn;
}tr[maxn<<2];

void pushup(int u){
	tr[u].minn=min(tr[u<<1].minn,tr[u<<1|1].minn);
}

void pushdown(int u){
	if(tr[u].laz){
		tr[u<<1].laz+=tr[u].laz;
		tr[u<<1].minn+=tr[u].laz;
		tr[u<<1|1].laz+=tr[u].laz;
		tr[u<<1|1].minn+=tr[u].laz;
		tr[u].laz=0;
	}
}

void build(int u,int l,int r){
	tr[u]={l,r,0,0};
	if(l==r) return ;
	int mid=(l+r)/2;
	build(u<<1,l,mid);build(u<<1|1,mid+1,r);
	pushup(u);
}

void update(int u,int l,int r,int val){
	if(tr[u].l>=l&&tr[u].r<=r){
		tr[u].minn+=val;
		tr[u].laz+=val;
		return ;
	}
	pushdown(u);
	int mid=(tr[u].l+tr[u].r)/2;
	if(l<=mid) update(u<<1,l,r,val);
	if(r>mid) update(u<<1|1,l,r,val);
	pushup(u);
}

ll query(int u,int l,int r){
	if(tr[u].l>=l&&tr[u].r<=r){
		return tr[u].minn;
	}
	pushdown(u);
	int mid=(tr[u].l+tr[u].r)/2;
	ll ans=1e18;
	if(l<=mid) ans=min(ans,query(u<<1,l,r));
	if(r>mid) ans=min(ans,query(u<<1|1,l,r));
	return ans;
}

int main(){
	n=read,m=read-1;
	rep(i,1,n){
		p[i].l=read,p[i].r=read-1,p[i].w=read;
	}	
	sort(p+1,p+1+n,cmp);
	build(1,1,m);
	int l=1,r=1;
	ll ans=1e18;
	update(1,p[1].l,p[1].r,1);
	while(r<n&&query(1,1,m)==0){
		r++;
		update(1,p[r].l,p[r].r,1);
	}
	while(query(1,1,m)>0){
		update(1,p[l].l,p[l].r,-1);
		l++;
	}
	ans=min(ans,p[r].w-p[l-1].w);
	while(r<n){
			while(r<n&&query(1,1,m)==0){
				r++;
				update(1,p[r].l,p[r].r,1);
			}	
			if(query(1,1,m)==0) break;
			while(query(1,1,m)>0){
				update(1,p[l].l,p[l].r,-1);
				l++;
			}
			ans=min(ans,p[r].w-p[l-1].w);
	}
	
	printf("%lld
",ans);
	
	return 0;
}