[题解]LuoGu3466：[POI2008]KLO-Building blocks

最新推荐文章于 2020-06-02 09:56:13 发布

原创最新推荐文章于 2020-06-02 09:56:13 发布 · 422 阅读

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#题解 #LuoGu #fhq Treap

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本文介绍了一种使用FHQ Treap数据结构解决寻找最优区间中位数的问题，通过枚举区间并利用Treap快速计算区间内各数值与中位数差的绝对值之和，实现O(n log n)的高效算法。

原题传送门
转化题意：
选一段长度为k的连续区间，使得区间内每个数减去中位数的差的绝对值之和最小

当然是枚举区间啦，然后 $O (l o g n)$ 求出当前区间每个数减去中位数的差的绝对值之和
工具？fhq Treap! pushup里面加维护一个sum就好啦

Code：

#include <bits/stdc++.h>
#define maxn 100010
#define LL long long
using namespace std;
int rt, sz, n, k, val[maxn], key[maxn], size[maxn], son[maxn][2], a[maxn];
LL sum[maxn];

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

int addnode(int x){
	++sz;
	val[sz] = sum[sz] = x, key[sz] = rand() * rand(), size[sz] = 1;
	return sz;
}

void pushup(int x){
	if (x){
		size[x] = 1, sum[x] = val[x];
		if (son[x][0]) size[x] += size[son[x][0]], sum[x] += sum[son[x][0]];
		if (son[x][1]) size[x] += size[son[x][1]], sum[x] += sum[son[x][1]];
	}
}

void split(int now, int w, int &u, int &v){
	if (!now) u = v = 0; else{
		if (val[now] <= w) u = now, split(son[now][1], w, son[u][1], v); else
		v = now, split(son[now][0], w, u, son[v][0]);
		pushup(now);
	}
}

int merge(int u, int v){
	if (!u || !v) return u + v;
	if (key[u] >= key[v]){
		son[u][1] = merge(son[u][1], v);
		pushup(u);
		return u;
	} else{
		son[v][0] = merge(u, son[v][0]);
		pushup(v);
		return v;
	}
}

int kth(int now, int k){
	while (1){
		if (size[son[now][0]] >= k) now = son[now][0]; else
		if (size[son[now][0]] + 1 >= k) return now; else
		k -= size[son[now][0]] + 1, now = son[now][1];
	}
}

void insert(int w){
	int x, y, z;
	split(rt, w, x, y);
	rt = merge(merge(x, addnode(w)), y);
}

void del(int w){
	int x, y, z;
	split(rt, w, x, y);
	split(x, w - 1, x, z);
	rt = merge(merge(x, merge(son[z][0], son[z][1])), y);
}

int main(){
	srand(time(0));
	n = read(), k = read();
	for (int i = 1; i <= n; ++i) a[i] = read();
	for (int i = 1; i < k; ++i) insert(a[i]);
	LL ans = 1e12;
	int ansl, ansr, ansmid;
	for (int i = k; i <= n; ++i){
		insert(a[i]);
		LL w = val[kth(rt, (k + 1) >> 1)];
		int x, y;
		split(rt, w, x, y);
		LL Sum = w * size[x] - sum[x] + sum[y] - w * size[y];
		rt = merge(x, y);
		if (Sum < ans){
			ans = Sum, ansl = i - k + 1, ansr = i, ansmid = w;
		}
		del(a[i - k + 1]);
	}
	printf("%lld\n", ans);
	for (int i = ansl; i <= ansr; ++i) a[i] = ansmid;
	for (int i = 1; i <= n; ++i) printf("%d\n", a[i]);
	return 0;
}