Codeforces Round #222 (Div. 1) D. Developing Game 线段树有效区间合并

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 5e5+10, M = 2e5+20, mod = 1e9+7, inf = 2e9;

int n;
int tag[N*4],mx[N*4];
void push_up(int i) {
        mx[i] = max(mx[ls],mx[rs]);
}
void push_down(int i,int ll,int rr) {
    if(tag[i] != 0 && ll != rr) {
        tag[ls] += tag[i];
        tag[rs] += tag[i];
        mx[ls] += tag[i];
        mx[rs] += tag[i];
        tag[i] = 0;
    }
}
void update(int i,int ll,int rr,int l,int r,int v)
{
    push_down(i,ll,rr);
    if(l == ll && r == rr) {
        tag[i] += v;
        mx[i] += v;
        return ;
    }
    if(r <= mid) update(ls,ll,mid,l,r,v);
    else if(l > mid) update(rs,mid+1,rr,l,r,v);
    else {
        update(ls,ll,mid,l,mid,v);
        update(rs,mid+1,rr,mid+1,r,v);
    }
    push_up(i);
}

int query(int i,int ll,int rr,int x) {
    push_down(i,ll,rr);
    if(ll == rr) return ll;
    if(mx[ls] == x) return query(ls,ll,mid,x);
    else return query(rs,mid+1,rr,x);
    push_up(i);
}
struct ss{
        int l,r,h,in;
        ss(int l = 0, int r = 0, int h = 0,int in = 0) : l(l), r(r), h(h),in(in) {}
        bool operator < (const ss & b) const {
            return h < b.h || h == b.h && in > b.in;
        }
}p[N],P[N];
int main() {
        scanf("%d",&n);
        for(int i = 1; i <= n; ++i) {
            int l,v,r;
            scanf("%d%d%d",&l,&v,&r);
            p[i] = ss(l,v,v,1);
            p[i+n] = ss(l,v,r,-1);
            P[i] = ss(l,r,v,1);
        }
        int m = n << 1;
        sort(p+1,p+m+1);
        int ans = 0,x,y;
        for(int i = 1; i <= m; ++i) {
            int l = p[i].l, r = p[i].r;
            update(1,1,300000,l,r,p[i].in);
            if(mx[1] > ans) {
                ans = mx[1];
                x = query(1,1,300000,mx[1]);
                y = p[i].h;
            }
        }
        printf("%d\n",ans);
        for(int i = 1; i <= n; ++i) {
            if(P[i].l <= x && P[i].r >= y && P[i].h >= x && P[i].h <= y) printf("%d\n",i);
        }
    return 0;
}