[Bzoj1112][POI2008]砖块Klo（splay）

1112: [POI2008]砖块Klo

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Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 2353  Solved: 831
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Description

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N柱砖,希望有连续K柱的高度是一样的. 你可以选择以下两个动作 1:从某柱砖的顶端拿一块砖出来,丢掉不要了. 2:从仓库中拿出一块砖,放到另一柱.仓库无限大. 现在希望用最小次数的动作完成任务.

Input

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第一行给出N,K. (1 ≤ k ≤ n ≤ 100000), 下面N行,每行代表这柱砖的高度.0 ≤ hi ≤ 1000000

Output

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最小的动作次数

Sample Input

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Sample Output

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HINT

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原题还要求输出结束状态时,每柱砖的高度.本题略去.

分析：

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splay求中位数。

AC代码：

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# include <iostream>
# include <cstdio>
# include <cstring>
using namespace std;
typedef long long LL;
const int N = 1e5 + ;
LL s[N],ret = 1e18;int n,K,root,cnt;
struct node{
   int ch[];
   int fa;
   int sz;LL val,sum;
   node(){}
   node(LL val):fa(),sz(),val(val),sum(sum){ch[] = ch[] = ;}
}a[N];
void push(int x)
{
    a[x].sz = a[a[x].ch[]].sz + a[a[x].ch[]].sz + ;
    a[x].sum = a[a[x].ch[]].sum + a[a[x].ch[]].sum + a[x].val;
}
void rotate(int now,int d)
{
    int pre = a[now].fa,g = a[pre].fa,nex = a[now].ch[d];
    a[pre].ch[d ^ ] = nex;
    if(nex)a[nex].fa = pre;
    a[now].fa = g;
    if(g)a[g].ch[a[g].ch[] == pre] = now;
    a[pre].fa = now;a[now].ch[d] = pre;
    push(pre);push(now);
}
void splay(int now)
{
    int pre,g;
    while(a[now].fa)
    {
        pre = a[now].fa,g = a[pre].fa;
        if(g && !((a[pre].ch[] == now) ^ (a[g].ch[] == pre)))rotate(pre,a[pre].ch[] == now);
        rotate(now,a[pre].ch[] == now);
    }
    root = now;
}
void insert(LL val)
{
    if(!root){a[++cnt] = node(val);root = cnt;return;}
    int now = root,pre;
    while(now)
    {
        pre = now;
        now = a[now].ch[val > a[now].val];
    }
    a[++cnt] = node(val);a[cnt].fa = pre;
    a[pre].ch[val > a[pre].val] = cnt;
    splay(cnt);
}
void erase(int now)
{
    splay(now);
    if(!a[now].ch[] || !a[now].ch[]){root = a[now].ch[] + a[now].ch[];}
    else
    {
        int p = a[now].ch[];
        while(a[p].ch[])p = a[p].ch[];
        a[p].ch[] = a[now].ch[];
        a[a[now].ch[]].fa = p;
        root = a[now].ch[];a[root].fa = ;
        splay(a[now].ch[]);
    }
    a[root].fa = ;
}
int find(int k)
{
    int now = root;
    while(k > )
    {
        if(a[a[now].ch[]].sz < k){k -= a[a[now].ch[]].sz;k--;if(k > )now = a[now].ch[];}
        else now = a[now].ch[];
    }
    return now;
}
int main()
{
  scanf("%d %d",&n,&K);
  for(int i = ;i <= n;i++)scanf("%lld",&s[i]);
  for(int i = ;i < K;i++)insert(s[i]);
  for(int i = K;i <= n;i++)
  {
      if(i - K)erase(i - K);
      insert(s[i]);
      int now = find((K + ) >> );
      splay(now);
      LL s1 = a[a[now].ch[]].sz,s2 = a[a[now].ch[]].sz;
      LL pre = a[a[now].ch[]].sum,bef = a[a[now].ch[]].sum;
      if(s1 * a[now].val - pre + bef - s2 * a[now].val < ret)ret = s1 * a[now].val - pre + bef - s2 * a[now].val;
  }
  printf("%lld\n",ret);
}