[BZOJ1150][CTSC2007]数据备份Backup

首先肯定要选择相邻的公司配对才会距离最小

先把两两之间的距离加入小根堆中

每次取出堆顶，加入答案，删掉左右线段，再扔进去len[l]+len[r]-len[now]

这样如果这个新点被取到，说明取两边而不取中间

做K次，贪心就是正确的了

/**************************************************************
    Problem: 1150
    User: momoka
    Language: C++
    Result: Accepted
    Time:620 ms
    Memory:6000 kb
****************************************************************/
 
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#define rep(i,s,t) for(int i=s;i<=t;++i)
#define drep(i,t,s) for(int i=t;i>=s;--i)
#define inf 0x3f3f3f3f
using namespace std;
long long ans,val[150010];
int l[150010],r[150010],id,n,k,d[100001];
typedef pair<long long,int>ele;
priority_queue<ele,vector<ele>,greater<ele> >q;
bool mark[150010];
inline int read(){
    char ch=getchar();
    int f=1,x=0;
    while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}
int main(){
   n=read(); k=read();
   rep(i,1,n) d[i]=read();
   rep(i,1,n-1){
     val[++id]=d[i+1]-d[i];
     q.push(make_pair(val[id],id));
     l[id]=id-1;
     r[id]=id+1;
   }
   r[0]=1; val[0]=1LL<<61; //q.push(make_pair(val[0],0));
   l[id+1]=id; val[id+1]=1LL<<61; // q.push(make_pair(val[id+1],id+1));
   ++id;
   rep(i,1,k){
     while (mark[q.top().second]) q.pop();
     ele now=q.top();
     q.pop();
     int idn=now.second;
     ans+=now.first;
     int ls=l[idn],rs=r[idn];
     mark[ls]=1;
     mark[rs]=1;
     l[idn]=l[ls]; r[l[idn]]=idn;
     r[idn]=r[rs]; l[r[idn]]=idn;
     val[idn]=val[ls]+val[rs]-now.first;
     q.push(make_pair(val[idn],idn));
   }
  printf("%lld\n",ans);
  return 0;
}