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

Posted 2020-08-16 Meek

D. Developing Game

 

Pavel is going to make a game of his dream. However, he knows that he can‘t make it on his own so he founded a development company and hired n workers of staff. Now he wants to pick n workers from the staff who will be directly responsible for developing a game.

Each worker has a certain skill level vi. Besides, each worker doesn‘t want to work with the one whose skill is very different. In other words, the i-th worker won‘t work with those whose skill is less than li, and with those whose skill is more than ri.

Pavel understands that the game of his dream isn‘t too hard to develop, so the worker with any skill will be equally useful. That‘s why he wants to pick a team of the maximum possible size. Help him pick such team.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of workers Pavel hired.

Each of the following n lines contains three space-separated integers li, vi, ri (1 ≤ li ≤ vi ≤ ri ≤ 3·105) — the minimum skill value of the workers that the i-th worker can work with, the i-th worker‘s skill and the maximum skill value of the workers that the i-th worker can work with.

Output

In the first line print a single integer m — the number of workers Pavel must pick for developing the game.

In the next line print m space-separated integers — the numbers of the workers in any order.

If there are multiple optimal solutions, print any of them.

Examples

input

4
2 8 9
1 4 7
3 6 8
5 8 10

output

3
1 3 4

 题意：

　　给你n个人的 位置v[i],同时每个人有一个可接受范围 l[i], r[i];

　　现在让你选尽量多的人，使得被选的人 都能互相接受

　　输出人数及选择方案

题解：

　　这个题看到不太会

　　对于两个人来说要让它们相互接受可以有以下情况

　　　　　　　　　　l1          v1              r1

　　　　　　　　          l2  v2　　    r2

　　　　　　　　　　　  l2            v2     r2　

　　　　　　　　　　　 l2           v2              r2

　　　　　　　　　　　l2                     v2     r2

　　发现一些规则

　　就是相交范围就是l1 v1 与 l2 v2的相交的一段，但是当v2超过r1的时候，就不可行了

　　所有我们可以想到一种 加上，减去的操作，

　　当v1 出现我们在线段树上加入l1,v1的这段有效的区间

　　当走过r1的时候 把l1 v1在线段树上减去即可 

　　显然对于一条线段要拆成两条线段, l1 v1 1 v1 和   l1 v1 -1 v2

　　再在线段树上操作就可以了

　　统计答案的时候呢，也可以利用线段树有效区间出来

　　@doubility

#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;
}