正题

P6320

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题目大意

有n个点，相邻的点不能同时选，问你选k个的最小代价

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解题思路

考虑贪心取每个点，取了一个点后设置一个撤回点，就是把该点的选择取反，同时选择左右，这样直接用堆维护即可

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code

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 100100
#define mp make_pair
#define fs first
#define sn second
using namespace std;
int n,k,x,lst,ans,l[N],r[N],s[N],p[N];
priority_queue<pair<int,int> >d;
int main()
{scanf("%d%d%d",&n,&k,&lst);for(int i=1;i<n;++i){scanf("%d",&x);s[i]=x-lst;lst=x;l[i]=i-1;r[i]=i+1;d.push(mp(-s[i],i));}s[0]=s[n]=1e9; while(k--){while(p[d.top().sn])d.pop();x=d.top().sn;d.pop();ans+=s[x];s[x]=s[l[x]]+s[r[x]]-s[x];p[l[x]]=1;p[r[x]]=1;l[x]=l[l[x]];r[x]=r[r[x]];r[l[x]]=x;l[r[x]]=x;d.push(mp(-s[x],x));}printf("%d",ans);return 0;
}