解题：APIO/CTSC 2007 数据备份

Posted 2021-01-13 Speranza_Leaf

题面

用双向链表把相邻两项的差串起来，用大根堆维护价值，每次贪心取最大的$x$。取完之后打标记删掉$pre[x]$和$nxt[x]$，之后用$val[pre[x]]+val[nxt[x]]-val[x]$替换这个$x$塞进堆里去，注意边界要连上一个极值

 1 #include<queue> 
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 const int N=100005;
 7 struct a{long long val,pos;};
 8 bool operator < (a x,a y)
 9 {
10     return x.val>y.val;
11 }
12 priority_queue<a> hp;
13 int pre[N],nxt[N],del[N];
14 long long n,m,rd,last,ans,b[N];
15 void Delete(int pos)
16 {
17     del[pos]=true;
18     pre[nxt[pos]]=pre[pos];
19     nxt[pre[pos]]=nxt[pos];
20 }
21 int main ()
22 {
23     scanf("%lld%lld",&n,&m),n--,b[0]=1e12;
24     pre[1]=0,nxt[1]=2,pre[n]=n-1,nxt[n]=0;    
25     for(int i=2;i<n;i++)
26         pre[i]=i-1,nxt[i]=i+1;
27     scanf("%lld",&last); 
28     for(int i=1;i<=n;i++)
29     {
30         scanf("%lld",&rd),b[i]=rd-last; 
31         hp.push((a){b[i],i}),last=rd;
32     }
33     while(m--)
34     {
35         while(del[hp.top().pos]) hp.pop();
36         int p=hp.top().pos; hp.pop();
37         ans+=b[p],b[p]=b[pre[p]]+b[nxt[p]]-b[p];
38         hp.push((a){b[p],p}),Delete(pre[p]),Delete(nxt[p]);
39     }
40     printf("%lld",ans);
41     return 0;
42 }

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