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Codeforces Round #222 (Div. 1) D. Developing Game 线段树有效区间合并
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.
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.
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.
4
2 8 9
1 4 7
3 6 8
5 8 10
3
1 3 4
#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_pairtypedef 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;}
Codeforces Round #222 (Div. 1) D. Developing Game 线段树有效区间合并