377D Developing Game(线段树+扫描线)

 

D. Developing Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 v_i. 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 l_i, and with those whose skill is more than r_i.

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 l_i, v_i, r_i (1 ≤ l_i ≤ v_i ≤ r_i ≤ 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.

Sample test(s)

input

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

output

3
1 3 4

input

6
3 5 16
1 6 11
4 8 12
7 9 16
2 10 14
8 13 15

output

4
1 2 3 5

题意：有n个开发人员，每个人有个技能值vi，每个人不想和技能值不在区间[li,ri]的人合作，现在要选尽可能多的人，使得满足所有选中的人的要求。

 

思路：首先，如果我们能求出一组L，R，使得每个选中的人i都满足li<=L<=R<=ri并且L<=vi<=R，那么我们就能找到应该选出的人。那么如何求这个L和R呢，把L，R想象成坐标系中的一个点，那么对于每一个人，需要满足li<=L<=vi并且vi<=R<=ri，这样，每个人可以看成一个矩形，问题就变成了寻找一个点使得有最多的矩形覆盖，这个问题可以用线段树+扫描线处理。

 

代码：

 

#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , mid
#define rson rs , mid + 1 , r
#define root 1 , 1 , n
#define rt o , l , r

vector <int> v1[300005],v2[300005];
int L[100005],V[100005],R[100005];
int N,ans;
int a,b;

struct node
{
    int x,w,c;//c表示子节点是否需要更新,0表示不用,
    //w存的是当前节点的最左端,x表示最大的覆盖数
}s[1<<20];

void build (int o , int l , int r)
{
    s[o].c = s[o].x = 0;
    if (l == r)
    {
        s[o].w = l;
        return;
    }
    int mid = ( l + r ) >> 1;
    build(lson);
    build(rson);
    s[o].w = s[ls].w;
}
void pushdown(int o)
{
    s[ls].x += s[o].c;
    s[rs].x += s[o].c;
    s[ls].c += s[o].c;
    s[rs].c += s[o].c;
    s[o].c = 0;
}
void update(int o , int l , int r , int le , int re , int v)
{
    if(l <= le && re <= r)
    {
        s[o].x += v;
        s[o].c += v;
        return ;
    }
    if(s[o].c != 0) pushdown(o);
    int mid = ( le + re ) >> 1;
    if(r > mid)
        update(rs, l, r , mid+1 , re , v);
    if(l <= mid)
        update(ls, l, r , le , mid , v);

    if(s[ls].x < s[rs].x)
    {
        s[o].x = s[rs].x;
        s[o].w = s[rs].w;
    }
    else
    {
        s[o].x = s[ls].x;
        s[o].w = s[ls].w;
    }
}
int main()
{
    scanf("%d",&N);
    for(int i=0; i<N; i++)
    {
        scanf("%d%d%d",L+i,V+i,R+i);
        v1[L[i]].push_back(i);
        v2[V[i]].push_back(i);
    }
    build(1,1,3e5);
    for(int i=1; i<=3e5; i++)
    {
        for(int j=0; j<v1[i].size(); j++)
        {
            update(1,V[v1[i][j]],R[v1[i][j]],1,3e5,1);
            //printf("i=%d , v1=%d , V=%d , R=%d\n",i,v1[i][j],V[v1[i][j]],R[v1[i][j]]);
        }
        if(ans<s[1].x)
        {
            ans=s[1].x, a=i, b=s[1].w;
            //printf("ans=%d , a=%d , b=%d\n",ans,a,b);
        }
        for(int j=0; j<v2[i].size(); j++)
        {
            update(1,i,R[v2[i][j]],1,3e5,-1);
            //printf("i=%d , v2=%d , i=%d , R=%d\n",i,v2[i][j],V[v2[i][j]],R[v2[i][j]]);
        }
    }
    printf("%d\n",ans);
    for(int i=0; i<N; i++)
    {
        if(L[i]<=a&&a<=V[i]&&V[i]<=b&&b<=R[i])
            printf("%d ",i+1);
    }
    return 0;
}