BZOJ 1112: [POI2008]砖块Klo1112( BST )

枚举每个长度为k的区间, 然后用平衡树找中位数进行判断, 时间复杂度O(nlogn).

早上起来精神状态不太好...连平衡树都不太会写了...果断去看了会儿番然后就A了哈哈哈

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#include<bits/stdc++.h>

  

#define rep(i, n) for(int i = 0; i < n; i++)

#define clr(x, c) memset(x, c, sizeof(x))

 

using namespace std;

 

typedef long long ll;

 

const int maxn = 100009;

const int inf = 10000000;

 

int n, k;

 

struct Node {

Node *ch[2], *p;

int s, v;

ll sum;

inline void setc(Node* t, int d) {

ch[d] = t;

t->p = this;

}

inline bool d() {

return this == p->ch[1];

}

inline void upd() {

s = ch[0]->s + ch[1]->s + 1;

sum = ch[0]->sum + ch[1]->sum + v;

}

} pool[maxn], *pt = pool, *root, *null;

 

Node* newNode(int v) {

pt->p = pt->ch[0] = pt->ch[1] = null;

pt->s = 1;

pt->v = pt->sum = v;

return pt++;

}

 

void init() {

null = pt++;

null->s = null->sum = null->v = 0;

null->ch[0] = null->ch[1] = null->p = null;

root = newNode(inf);

root->setc(newNode(-inf), 0);

root->upd();

}

 

void rot(Node* t) {

Node* p = t->p;

int d = t->d();

p->p->setc(t, p->d());

p->setc(t->ch[d ^ 1], d);

t->setc(p, d ^ 1);

p->upd();

if(root == p) root = t;

}

 

void splay(Node* t, Node* f = null) {

while(t->p != f) {

if(t->p->p != f)

   t->d() != t->p->d() ? rot(t) : rot(t->p);

rot(t);

}

t->upd();

}

 

Node* select(int k) {

for(Node* t = root; ;) {

int s = t->ch[0]->s;

if(k == s) return t;

if(k > s)

   t = t->ch[1], k -= s + 1;

else

   t = t->ch[0];

}

}

 

int rank(int v) {

int ans = 0;

for(Node* t = root; t != null;) {

if(t->v < v) 

   ans += t->ch[0]->s + 1, t = t->ch[1];

else

   t = t->ch[0];

}

return ans;

}

 

void insert(int v) {

int r = rank(v);

Node *L = select(r - 1), *R = select(r);

splay(L), splay(R, L);

R->setc(newNode(v), 0);

R->upd(), L->upd();

}

 

void del(int v) {

int r = rank(v);

Node *L = select(r - 1), *R = select(r + 1);

splay(L), splay(R, L);

R->setc(null, 0);

R->upd(), L->upd();

}

 

ll work() {

Node *L = select(0), *R = select(k + 1), *t = select((k >> 1) + 1);

splay(L), splay(R, L);

splay(t, R);

return t->v * (t->ch[0]->s - t->ch[1]->s) + t->ch[1]->sum - t->ch[0]->sum;

}

 

int A[maxn];

 

int main() {

freopen("test.in", "r", stdin);

freopen("test.out", "w", stdout);

init();

ll ans;

cin >> n >> k;

rep(i, n) {

   scanf("%d", A + i);

   if(i < k) insert(A[i]);

}

ans = work();

for(int i = k; i < n; i++) {

insert(A[i]);

del(A[i - k]);

ans = min(ans, work());

}

cout << ans << "\n";

return 0;

}

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1112: [POI2008]砖块Klo

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 1288  Solved: 444
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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
3
9
2
3
1

Sample Output

2

HINT

原题还要求输出结束状态时,每柱砖的高度.本题略去.

Source