CF1295E Permutation Separation

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纪念一道傻逼题。

排列$p_i$,从中分割2堆，开始移动元素使得左边元素小于右边，移动代价为$a_i$

移动结束后代价变成$1-x$,$x+1-n$。这样代价为$ans_x$答案就是$\sum_{i=0}^{n} ans_x$。假设都在右边。

枚举分割点$i$,$x<=p[i]，p[i]一开始在左边，就需要移动代价为a_i$,$x>p[i]，p[i]不需要移动，就多余的移动-a_i$

代码 
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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
const int mod = 1e9 + 7;
struct node
{
    int l, r;
    ll minn, lazy;
} tree[N << 2];
void pushup(int pos)
{
    tree[pos].minn = min(tree[pos * 2].minn, tree[2 * pos + 1].minn);
}
void addlazy(int pos,ll w)
{
    tree[pos].lazy += w;
    tree[pos].minn += w;
}
void pushdown(int pos)
{
    if (tree[pos].lazy)
    {
        addlazy(pos<<1,tree[pos].lazy);
        addlazy(pos<<1|1,tree[pos].lazy);
        tree[pos].lazy=0;
    }
}
void build(int pos, int l, int r)
{
    tree[pos].l = l;
    tree[pos].r = r;
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    build(pos <<1, l, mid);
    build(pos <<1 |1, mid + 1, r);
}
void update(int pos, int l, int r,ll w)
{
    if (l <= tree[pos].l && tree[pos].r <= r)
    {
        addlazy(pos,w);
        return;
    }
    pushdown(pos);
    int mid = (tree[pos].l + tree[pos].r) >> 1;
    if (l <= mid)
        update(pos<<1, l, r, w);
    if (r > mid)
        update(pos<<1|1, l, r, w);
    pushup(pos);
}
int p[N];
ll a[N];
int n;
ll ans = 1e18;
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &p[i]);
    for (int i = 1; i <= n; i++)
        scanf("%lld", &a[i]);
    build(1,0,n);
    for (int i = 1; i <= n; i++)
        update(1, p[i] , n, a[i]); 
   
    for (int i = 1; i < n; i++)
    {
        update(1, p[i] , n, -a[i]);
        update(1, 0, p[i]-1, a[i]); 
       
        ans = min(ans, tree[1].minn);
    }
    printf("%lld\n", ans);
}