BZOJ 5102: [POI2018]Prawnicy

考虑最优解的集合中一定有一个$l$最大的，我们就去枚举左端点，把所有$l$小于等于它的全丢进堆里，取前$k$个即可。

#include <bits/stdc++.h> 
using namespace std;

#define N 1000010
#define fi first
#define se second
#define pii pair <int, int>
int n, k;
struct node
{
    int l, r, id;
    void scan(int _id)
    {
        id = _id;
        scanf("%d%d", &l, &r);
    }
    bool operator < (const node &other) const
    {
        return l < other.l;
    }
}a[N];

void work()
{
    priority_queue <pii, vector <pii>, greater <pii> > pq; 
    int Max = 0, L = -1, R = -1;
    for (int i = 1; i <= n; ++i)
    {
        pq.push(pii(a[i].r, a[i].id));
        if ((int)pq.size() > k)
            pq.pop(); 
        if ((int)pq.size() == k)
        {
            int t = max(0, pq.top().first - a[i].l);
            if (t > Max)
            {
                Max = t;
                L = a[i].l;
                R = pq.top().fi;
            }
        }
    }
    printf("%d
", Max);
    for (int i = 1; i <= n; ++i)
    {
        if (a[i].l <= L && a[i].r >= R)
        {
            printf("%d", a[i].id);
            if (k == 1)
            {
                puts("");
                return;
            }
            else
            {
                --k;
                putchar(' ');
            }
        }
    }
}

void Run()
{
    while (scanf("%d%d", &n, &k) != EOF)
    {
        for (int i = 1; i <= n; ++i)
            a[i].scan(i);
        sort(a + 1, a + 1 + n);
        work();
    }
}

int main()
{
    #ifdef LOCAL
        freopen("Test.in", "r", stdin);
    #endif 

    Run();
    return 0;
}