POI2008 KLO-Building blocks
死因:不认真读题
首先,我们可以想到暴力枚举每段长度为\(k\)的区间,对于这一段区间,求将它变为同一高度的最小操作次数。显然,当我们取区间的中位数时,这段区间变为同一高度的次数最小。
操作次数为:
对于大于中位数的数,求和,减去中位数乘它们的个数
对于小于等于中位数的数,中位数乘它们的个数减去它们的和
最后两项求和即可
所以我们需要支持的操作有:
- 求这段区间的中位数
- 求小于中位数的数的个数
- 求小于中位数的数的和
可以使用平衡树,在这里我是用的是权值线段树,比较方便。
最后要注意的是,高度可以为0
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define LL long long
#define ls p << 1
#define rs p << 1 | 1
#define mid ((l + r) >> 1)
using namespace std;
LL read() {
LL k = 0, f = 1; char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9')
k = k * 10 + c - 48, c = getchar();
return k * f;
}
LL a[100010], n, k, maxn, sum[100010];
struct zzz {
LL cnt[1000010 << 2]; LL sum[1000010 << 2];
void up(LL p) {
cnt[p] = cnt[ls] + cnt[rs];
sum[p] = sum[ls] + sum[rs];
}
void insert(LL x, LL p = 1, LL l = 0, LL r = maxn) {
if(l == r) {
++cnt[p]; sum[p] += l;
//cout << cnt[p] << ' ' << l << endl;
return ;
}
if(x <= mid) insert(x, ls, l, mid);
else insert(x, rs, mid+1, r);
up(p);
}
void earse(LL x, LL p = 1, LL l = 0, LL r = maxn) {
if(l == r) {
--cnt[p]; sum[p] -= l; return ;
}
if(x <= mid) earse(x, ls, l, mid);
else earse(x, rs, mid+1, r);
up(p);
}
LL k_th(LL x, LL p = 1, LL l = 0, LL r = maxn) {
if(l == r) return l;
if(x <= cnt[ls]) return k_th(x, ls, l, mid);
else return k_th(x - cnt[ls], rs, mid+1, r);
}
LL summ(LL nr, LL p = 1, LL l = 0, LL r = maxn, LL nl = 0) {
LL ans = 0;
if(nl <= l && nr >= r) return sum[p];
if(nl <= mid) ans += summ(nr, ls, l, mid, nl);
if(nr > mid) ans += summ(nr, rs, mid+1, r, nl);
return ans;
}
LL numm(LL nr, LL p = 1, LL l = 0, LL r = maxn, LL nl = 0) {
LL ans = 0;
if(nl <= l && nr >= r) return cnt[p];
if(nl <= mid) ans += numm(nr, ls, l, mid, nl);
if(nr > mid) ans += numm(nr, rs, mid+1, r, nl);
return ans;
}
}tree;
int main() {
n = read(), k = read();
LL anss = -1, pos, num;
for(LL i = 1; i <= n; ++i) a[i] = read(), maxn = max(a[i], maxn), sum[i] = a[i] + sum[i-1];
for(LL i = 1; i <= n; ++i) {
tree.insert(a[i]);
if(i > k) tree.earse(a[i-k]);
if(i >= k) {
LL x = tree.k_th((k+1) >> 1);
LL y = tree.summ(x), summ = sum[i] - sum[i-k], numl = tree.numm(x);
//cout << y << ' ' << numl << ' ' << summ << endl;
if(summ-y-x*(k-numl)+x*numl-y < anss || anss == -1) {
anss = summ-y-x*(k-numl)+x*numl-y, pos = i; num = x;
}
}
}
printf("%lld\n", anss);
for(LL i = 1; i <= n; ++i) {
if(i >= pos-k+1 && i <= pos) {
printf("%lld\n", num);
}
else printf("%lld\n", a[i]);
}
return 0;
}