Luo Gu [P1484] [trees] [greedy] heap

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Subject to the effect:

Topic links: https://www.luogu.org/problemnew/show/P1484
have $n$ points, wherein select adjacent to each other does not exceed $m$ points, so that the right and the maximum point.

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Ideas:

Thinking problems.
Consider greedy. If we choose the maximum points for the first time $i$ When, then select two points or points $i-1$ sum point $i+1$ election together, or neither of the two.

If we choose to point $i-1$ sum point $x(x≠i+1)$ , then the greedy thoughts, $a[i-1]+a[x]$ must be greater than any of $a[q]+a[p]$ . But when $p=i,q=x$ time, $a[i-> 1]+a[x]$ is certainly less than $a[i]+a[x]$ (For a start, $a[i]$ is the greatest). contradiction.
QED

If you do not choose $a[i-1]$ and $a[i+1]$ , then it is a normal greedy. But if you choose $a[i-1]$ and $a[i+1]$ , then it needs to meet revocable.
If we $a[i]$ is changed to the value of $a[i-1]+a[i+1]-a[i]$ , selected again $a$ array maximum value, $years$ is equal to $a[i]+a[i-1]+a[i+1]-a[i]=a[i-1]+a[i+1]$！

So used to maintain the maximum heap, it can be done $O(m\log n)$ complexity

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Code:

#include <queue>
#include <cstdio>
#define mp make_pair
using namespace std;
typedef long long ll;

const int N=500010;
int n,m;
bool p[N];
ll a[N],ans;
priority_queue<pair<ll,int> > q;

struct node
{
	int l,r;
}link[N];

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		q.push(mp(a[i],i));
		link[i].l=i-1; link[i].r=i+1;
	}
	while (m--)
	{
		while (p[q.top().second]) q.pop();
		if (q.top().first<0) break;
		ans+=q.top().first;
		int id=q.top().second;
		a[id]=a[link[id].l]+a[link[id].r]-a[id];
		p[link[id].l]=p[link[id].r]=1;
		link[id].l=link[link[id].l].l; link[link[id].l].r=id;
		link[id].r=link[link[id].r].r; link[link[id].r].l=id;
		q.pop(); q.push(mp(a[id],id));
	}
	printf("%lld",ans);
	return 0;
}