BZOJ5102 POI2018Prawnicy（堆）

　　考虑固定右端点，使左端点最小。那么按右端点排序后查询前缀这些区间的左端点第k小即可。然而写了一个treap一个线段树都T飞了，感觉惨爆。事实上可以用堆求第k小，维护一个大根堆保证堆中元素不超过k个即可，瞬间就跑的飞快了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define ll long long
#define N 1000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,ans,u,v;
struct data
{
    int l,r,i;
    bool operator <(const data&a) const
    {
        return r<a.r;
    }
}a[N];
priority_queue<int> q;
void print(int x,int p)
{
    for (int i=p;i<=n;i++)
    if (a[i].l<=x) printf("%d ",a[i].i);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5102.in","r",stdin);
    freopen("bzoj5102.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    n=read(),m=read();
    for (int i=1;i<=n;i++) a[i].l=read(),a[i].r=read(),a[i].i=i;
    sort(a+1,a+n+1);
    for (int i=n;i>=1;i--)
    {
        q.push(a[i].l);
        if (n-i+1>=m)
        {
            int x=q.top();q.pop();
            if (a[i].r-x>ans) u=x,v=i,ans=a[i].r-x;
        }
    }
    cout<<ans<<endl;print(u,v);
    return 0;
}