[POI2008]砖块Klo

Time Limit: 10 Sec Memory Limit: 162 MB

Description

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

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

最小的动作次数

Sample Input

5 3

Sample Output

HINT

原题还要求输出结束状态时,每柱砖的高度.本题略去.

初中老师教我们肯定选中位数。。。
所以我们要维护这个东西。。。
那就平衡树呗。。。
然鹅我打板不是很6
一直TLE。。。用血的教训证明了 Splay 真的要看你和你的父亲是不是弯的啥的。。。
真的不能无脑转到根啊。。。


#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5, L = 0, R = 1;
const long long INF = 1e11 + 5;
struct lpl{
    int size, tim, fa, son[2];
    long long data, sum;
}node[maxn];
int n, k, lin, cnt, root, ini[maxn];
long long ans = INF, sum1, sum2;

inline void update(int t)
{
    node[t].size = node[node[t].son[L]].size + node[node[t].son[R]].size + node[t].tim;
    node[t].sum = node[node[t].son[L]].sum + node[node[t].son[R]].sum + node[t].data * node[t].tim; 
}

inline void rotate(int t)
{
    int fa = node[t].fa, grdfa = node[fa].fa, which = (node[fa].son[R] == t);
    node[grdfa].son[node[grdfa].son[R] == fa] = t; node[t].fa = grdfa;
    node[fa].son[which] = node[t].son[which ^ 1]; node[node[t].son[which ^ 1]].fa = fa;
    node[t].son[which ^ 1] = fa; node[fa].fa = t;
    update(fa); update(t);
}

inline void Splay(int t, int k)
{
    while(node[t].fa != k){
        int fa = node[t].fa, grdfa = node[fa].fa;
        if(grdfa != k){
            if((node[fa].son[R] == t) ^ (node[grdfa].son[R] == fa)) rotate(t);
            else rotate(fa);
        }
        rotate(t);
    }
    if(!k) root = t;
}

inline void insert(long long t)
{
    int now = root, fa = 0;
    while(1){
        if(!now) break;
        if(node[now].data == t) break;
        fa = now;
        if(t < node[now].data) now = node[now].son[L];
        else now = node[now].son[R];
    }
    if(now){
        node[now].size++; node[now].tim++; Splay(now, 0); return;
    }
    node[fa].son[t > node[fa].data] = ++cnt;
    node[cnt].fa = fa; node[cnt].data = t; node[cnt].tim = node[cnt].size = 1;
    Splay(cnt, 0);
}

inline void delet(int t)
{
    int now = root;
    while(t != node[now].data && node[now].son[t > node[now].data] != 0) now = node[now].son[t > node[now].data];
    Splay(now, 0);
    if(node[now].tim > 1){node[now].tim--; update(now); return;}
    if(!node[now].son[L]){root = node[now].son[R]; node[root].fa = 0; return;}
    int pre = node[root].son[L];
    while(node[pre].son[R] != 0) pre = node[pre].son[R];
    Splay(pre, root);
    node[node[root].son[R]].fa = pre; node[pre].fa = 0;  node[pre].son[R] = node[root].son[R]; 
    root = pre; update(root);       
}

void find(int t, int rank)
{   
    if(!t) return;
    if(rank > node[node[t].son[L]].size && rank <= node[node[t].son[L]].size + node[t].tim){
        sum1 += node[node[t].son[L]].sum + (rank - node[node[t].son[L]].size - 1) * node[t].data;
        sum2 += node[node[t].son[R]].sum + (node[node[t].son[L]].size + node[t].tim - rank) * node[t].data;
        lin = node[t].data;
    }
    else if(rank <= node[node[t].son[L]].size){
        sum2 += (node[node[t].son[R]].sum + node[t].tim * node[t].data);
        find(node[t].son[L], rank);
    }
    else{
        sum1 += (node[t].data * node[t].tim + node[node[t].son[L]].sum);
        find(node[t].son[R], rank - node[node[t].son[L]].size - node[t].tim);
    }
}

inline void workk()
{
    sum1 = sum2 = 0;
    int t = (k + 1) / 2;
    find(root, t); //sum1 += INF; sum2 -= INF;
    long long w = (t - 1) * lin - sum1 + sum2 - (k - t) * lin;
    ans = min(ans, w);
}

int main()
{
    //freopen("data.in", "r", stdin);
    //freopen("lpl.out", "w", stdout);
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; ++i) scanf("%d", &ini[i]);
    //insert(INF); insert(-INF);
    for(int i = 1; i < k; ++i) insert(ini[i]);
    for(int i = k; i <= n; ++i){
        insert(ini[i]); 
        workk(); 
        delet(ini[i - k + 1]);
    }
    cout << ans;
    return 0;
}

bzoj1112 [POI2008]砖块Klo