XV Olimpiada Informatyczna - Etap 1 - klo

题目大意:有一个长为 N 的整数数列,每次可以把一个数增减一,求最少次数使得有连续 K 个数相同。
下面程序用 Size balanced tree 实现,卫星数据是子树节点数和子树关键字和。

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#include <cstdio>
#include <climits>
#include <algorithm>
using namespace std;
const int N = 100000;
int a[N];
struct Node *null, *pit, *root;
struct Node
{
int key, size;
long long sum;
Node *ch[2];
Node() {}
Node(int key) : key(key), size(1), sum(key) {ch[0] = ch[1] = null; }
void update() {
size = ch[0]->size + 1 + ch[1]->size;
sum = ch[0]->sum + key + ch[1]->sum;
}
}pool[N+1];
void zag(Node *&x, int d)
{
Node *y = x->ch[!d];
x->ch[!d] = y->ch[d];
y->ch[d] = x;
y->size = x->size;
y->sum = x->sum;
x->update();
x = y;
}
void maintain(Node *&x, int d)
{
if (x == null) return;
if (x->ch[d]->ch[d]->size > x->ch[!d]->size)
zag(x, !d);
else if (x->ch[d]->ch[!d]->size > x->ch[!d]->size)
zag(x->ch[d], d), zag(x, !d);
else return;
maintain(x->ch[0], 0);
maintain(x->ch[1], 1);
maintain(x, 0);
maintain(x, 1);
}
void insert(Node *&x, int key)
{
if (x == null) {
x = new Node(key);
return;
}
int d = x->key < key;
++x->size;
x->sum += key;
insert(x->ch[d], key);
maintain(x, d);
}
Node *erase(Node *&x, int key)
{
if (x->key == key || key < x->key && x->ch[0] == null || key > x->key && x->ch[1] == null) {
if (x->ch[0] == null || x->ch[1] == null) {
Node *y = x;
x = x->ch[x->ch[0] == null];
return y;
}
Node *y = erase(x->ch[1], key-1);
x->key = y->key;
x->update();
return null;
}
Node *y = erase(x->ch[x->key < key], key);
x->update();
return y;
}
long long tot, mid;
void count(Node *&x, int cnt)
{
if (x->ch[0]->size == cnt) {
mid = x->key;
tot += x->ch[0]->sum + x->key;
} else if (x->ch[0]->size > cnt)
count(x->ch[0], cnt);
else
tot += x->ch[0]->sum + x->key,
count(x->ch[1], cnt-x->ch[0]->size-1);
}
int main()
{
pit = pool;
root = null = new Node(0); null->size = 0; null->ch[0] = null->ch[1] = null;
int n, k;
scanf("%d%d", &n, &k);
for (int i = 0; i < n; i++)
scanf("%d", &a[i]);
long long best = LLONG_MAX;
int res1, res2, res3;
for (int i = 0; i < k-1; i++)
insert(root, a[i]);
for (int i = k-1; i < n; i++) {
insert(root, a[i]);
tot = 0;
count(root, k/2);
long long t = mid*(k/2+1)-tot + root->sum-tot-mid*(k-k/2-1);
if (t < best) {
best = t;
res1 = i-k+1;
res2 = i;
res3 = mid;
}
erase(root, a[i-k+1]);
}
printf("%lld\n", best);
for (int i = res1; i <= res2; i++)
a[i] = res3;
for (int i = 0; i < n; i++)
printf("%d\n", a[i]);
}