BZOJ 1112。

题意相当于在一个长度为$k$的区间内选择一个数$s$使$\sum_{i = 1}^{k}\left | a_i - s \right |$最小。

很显然是中位数。

然后只要写一个能查询长度为$k$的区间的中位数,以及小于和大于这个中位数的总和和个数的数据结构即可。

线段树平衡树对顶堆随便维护。

我选择权值线段树。

时间复杂度$O(nlogn)$。

Luogu上还需要输出方案。

Code:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll; const int N = 1e5 + ; int n, K;
ll mn = 0LL, a[N]; template <typename T>
inline void read(T &X) {
X = ; char ch = ; T op = ;
for(; ch > '' || ch < ''; ch = getchar())
if(ch == '-') op = -;
for(; ch >= '' && ch <= ''; ch = getchar())
X = (X << ) + (X << ) + ch - ;
X *= op;
} template <typename T>
inline void chkMax(T &x, T y) {
if(y > x) x = y;
} namespace SegT {
struct Node {
int lc, rc;
ll sum, cnt;
} s[N * ]; int root, nodeCnt = ; #define lc(p) s[p].lc
#define rc(p) s[p].rc
#define sum(p) s[p].sum
#define cnt(p) s[p].cnt
#define mid ((l + r) >> 1) void ins(int &p, ll l, ll r, ll x) {
if(!p) p = ++nodeCnt;
sum(p) += x, ++cnt(p);
if(l == r) return; if(x <= mid) ins(lc(p), l, mid, x);
else ins(rc(p), mid + , r, x);
} void del(int &p, ll l, ll r, ll x) {
sum(p) -= x, --cnt(p);
if(l == r) return; if(x <= mid) del(lc(p), l, mid, x);
else del(rc(p), mid + , r, x);
} ll getKth(int p, ll l, ll r, int k) {
if(l == r) return l; int now = cnt(lc(p));
if(k <= now) return getKth(lc(p), l, mid, k);
else return getKth(rc(p), mid + , r, k - now);
} int qCnt(int p, ll l, ll r, ll x, ll y) {
if(x <= l && y >= r) return cnt(p); int res = ;
if(x <= mid) res += qCnt(lc(p), l, mid, x, y);
if(y > mid) res += qCnt(rc(p), mid + , r, x, y);
return res;
} ll qSum(int p, ll l, ll r, ll x, ll y) {
if(x <= l && y >= r) return sum(p); ll res = 0LL;
if(x <= mid) res += qSum(lc(p), l, mid, x, y);
if(y > mid) res += qSum(rc(p), mid + , r, x, y);
return res;
} #undef mid } using namespace SegT; int main() {
read(n), read(K);
for(int i = ; i <= n; i++) {
read(a[i]);
chkMax(mn, a[i]);
} // ll sum = 0LL;
for(int i = ; i <= K; i++) {
// sum += a[i];
ins(root, , mn, a[i]);
} int pos = ;
ll mid = getKth(root, , mn, (K + ) / );
ll minCost = mid * qCnt(root, , mn, , mid) - qSum(root, , mn, , mid);
minCost += qSum(root, , mn, mid + , mn) - mid * qCnt(root, , mn, mid + , mn);
for(int i = K + ; i <= n; i++) {
del(root, , mn, a[i - K]);
ins(root, , mn, a[i]); ll nowMid = getKth(root, , mn, (K + ) / );
ll nowCost = nowMid * qCnt(root, , mn, , nowMid) - qSum(root, , mn, , nowMid);
nowCost += qSum(root, , mn, nowMid + , mn) - nowMid * qCnt(root, , mn, nowMid + , mn); if(nowCost < minCost) {
pos = i - K + ;
mid = nowMid;
minCost = nowCost;
}
} printf("%lld\n", minCost);
/* for(int i = 1; i <= n; i++) {
if(i >= pos && i <= pos + K - 1) printf("%lld\n", mid);
else printf("%lld\n", a[i]);
} */ return ;
}

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