波动值之和

金牌导航 Splay-1

题目大意

给出一个数列，求${\sum }_{i=1}^{n}mi{n}_{j=1}^{i-1}|a_i-a_j|$

输入样例

6
5
1
2
5
4
6

输出样例

12

样例解释

$5+|1-5|+|2-1|+|5-5|+|4-5|+|6-5|=5+4+1+0+1+1=12$

数据范围

$1⩽n⩽32767,|a_i|⩽10^6$

解题思路

用Splay存起来，然后每次找到最接近当前数的数（把当前数旋转为根节点，然后在左子树找最大，在右子树找最小）

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 40000
using namespace std;
int n, x, w, rt, ans, s[N], ls[N], rs[N], fa[N];
int ss(int x)
{return x == ls[fa[x]];
}
void up(int x)
{int y = fa[x], z = fa[y], g = ss(x)? rs[x]: ls[x];if (z) ss(y)? ls[z] = x: rs[z] = x;if (ss(x)) rs[x] = y, ls[y] = g;else ls[x] = y, rs[y] = g;fa[x] = z;fa[y] = x;if (g) fa[g] = y;
}
void Splay(int x, int y)//旋转
{while(fa[x] != y){if (fa[fa[x]] != y)ss(x) == ss(fa[x])? up(fa[x]): up(x);up(x);}if (!y) rt = x;
}
void add(int v)//加点
{int x = rt, y = 0;while(x){y = x;if (v < s[x]) x = ls[x];else x = rs[x];}s[++w] = v;fa[w] = y;if (v < s[y]) ls[y] = w;else rs[y] = w;Splay(w, 0);
}
int ask()//查询
{int x = ls[rt], y = rs[rt], xx, yy;//分别查询左右子树if (x) while(rs[x]) x = rs[x];//左子树最大的if (y) while(ls[y]) y = ls[y];xx = s[rt] - s[x];yy = s[y] - s[rt];if (!x) return yy;if (!y) return xx;return min(xx, yy);
}
int main()
{scanf("%d", &n);scanf("%d", &ans);s[++w] = ans;rt = 1; while(--n){scanf("%d", &x);add(x);ans += ask();}printf("%d", ans);return 0;
}