解法参考黄源河左偏树论文
#include <stdio.h>
#include <iostream>
using namespace std;#define typec int // type of key val
#define N 50005int zabs(int x) {if (x < 0) return -x;else return x;
}const int na = -1;
struct node { typec key; int l, r, f, dist; } tr[N];int ary[N];
int tree[N], size[N], cunt[N];// 获得结点i的根
int iroot(int i) {if (i == na) return i;while (tr[i].f != na) i = tr[i].f;return i;
}// 合并两棵左偏树, param: rx ry (two root)
int merge(int rx, int ry) { if (rx == na) return ry;if (ry == na) return rx;if (tr[rx].key < tr[ry].key) swap(rx, ry);int r = merge(tr[rx].r, ry);tr[rx].r = r; tr[r].f = rx;if (tr[r].dist > tr[tr[rx].l].dist) swap(tr[rx].l, tr[rx].r);if (tr[rx].r == na) tr[rx].dist = 0;else tr[rx].dist = tr[tr[rx].r].dist + 1;return rx; // return new root
}// 插入一个新节点
int ins(int i, typec key, int root) {tr[i].key = key;tr[i].l = tr[i].r = tr[i].f = na;tr[i].dist = 0;return root = merge(root, i); // return new root
}// 删除某个结点
int del(int i) {if (i == na) return i;int x, y, l, r;l = tr[i].l; r = tr[i].r; y = tr[i].f;tr[i].l = tr[i].r = tr[i].f = na;tr[x = merge(l, r)].f = y;if (y != na && tr[y].l == i) tr[y].l = x;if (y != na && tr[y].r == i) tr[y].r = x;for ( ; y != na; x = y, y = tr[y].f) {if (tr[tr[y].l].dist < tr[tr[y].r].dist)swap(tr[y].l, tr[y].r);if (tr[tr[y].r].dist + 1 == tr[y].dist) break;tr[y].dist = tr[tr[y].r].dist + 1;}if (x != na) return iroot(x); // return new rootelse return iroot(y);
}// 获取最小结点
node top(int root) {return tr[root];
}// 取得并删除最小结点
node pop(int &root) {node out = tr[root];int l = tr[root].l, r = tr[root].r;tr[root].l = tr[root].r = tr[root].f = na;tr[l].f = tr[r].f = na;root = merge(l, r);return out;
}// 增/减一个结点的键值
int add(int i, typec val) {if (i == na) return i;if (tr[i].l == na && tr[i].r == na && tr[i].f == na) {tr[i].key += val;return i;}typec key = tr[i].key + val;int rt = del(i);return ins(i, key, rt);
}// 初始化数据
void init(int n) {for (int i = 0; i < n; i++) {scanf("%d", &ary[i]);tr[i].key = ary[i];tr[i].l = tr[i].r = tr[i].f = na;tr[i].dist = 0;}
}void solve(int n) {int m = -1;for (int i = 0; i < n; i++) {tree[++m] = i;size[m] = 1;cunt[m] = 1;while (m > 0 && top(tree[m]).key <= top(tree[m-1]).key) {tree[m-1] = merge(tree[m], tree[m-1]);size[m-1] += size[m];cunt[m-1] += cunt[m];m--;while (cunt[m] > (size[m] + 1) / 2) {pop(tree[m]);cunt[m]--;}}}int k = 0;long long res = 0;for (int i = 0; i <= m; i++) {int v = top(tree[i]).key;for (int j = 0; j < size[i]; j++) {res += zabs(ary[k++] - v);}}printf("%lld\n", res);
}int main() {int n;while (scanf("%d", &n), n) {init(n);solve(n);}return 0;
}