Solution -「CodeChef JUMP」Jump Mission

(mathcal{Description})

  Link.

  有 (n) 个编号 (1sim n) 的格子排成一排，并有三个权值序列 ({a_n},{h_n},{p_n})，其中 ({p_n}) 是一个排列。从 (i) 跳到 (j)，必须满足 (i<jland p_i<p_j)，代价为 ((h_i-h_j)^2+a_j)，求从 (1) 跳到 (n) 的最小代价。

  (n,h_ile6 imes10^5)。

(mathcal{Solution})

  不就是个板子套板子吗你还水题解。

  设 (f(i)) 表示从 (1) 跳到 (i) 的最小代价，显然

[egin{aligned}f(i) &= min_{j<i,p_j<p_i}{f(j)+a_i+(h_i-h_j)^2}\&= a_i+h_i^2+min_{j<i,p_j<p_i}{f(j)+h_j^2-2h_ih_j}.end{aligned} ]

转移条件是二维偏序关系，反手一个 CDQ，转移最优化的是一次函数，再丢一个李超树，复杂度 (mathcal O(nlog nlog h))。

  比树套树好写不知道多少，而且跑得飞快。

(mathcal{Code})

/*~Rainybunny~*/

#include <bits/stdc++.h>

#define rep( i, l, r ) for ( int i = l, rep##i = r; i <= rep##i; ++i )
#define per( i, r, l ) for ( int i = r, per##i = l; i >= per##i; --i )

inline char fgc() {
    static char buf[1 << 17], *p = buf, *q = buf;
    return p == q && ( q = buf + fread( p = buf, 1, 1 << 17, stdin ), p == q )
      ? EOF : *p++;
}

inline int rint() {
    int x = 0, f = 1, s = fgc();
    for ( ; s < '0' || '9' < s; s = fgc() ) f = s == '-' ? -f : f;
    for ( ; '0' <= s && s <= '9'; s = fgc() ) x = x * 10 + ( s ^ '0' );
    return x * f;
}

typedef long long LL;

inline void chkmin( LL& a, const LL b ) { b < a && ( a = b ); }

const int MAXN = 6e5;
const LL LINF = 1ll << 60;
int n, mxh, h[MAXN + 5], p[MAXN + 5], a[MAXN + 5];
LL f[MAXN + 5];
std::vector<int> ord[MAXN * 2 + 5];

struct Line {
    LL k, b;
    inline LL operator () ( const int x ) { return k * x + b; }
};

struct SegmentTree {
    Line line[MAXN + 5];
    int lcnt, vers, ver[MAXN << 2], idx[MAXN << 2];

    inline void clear() { lcnt = 0, ++vers; }

    inline void insert( const int u, const int l, const int r, int id ) {
        if ( ver[u] != vers ) return idx[u] = id, ver[u] = vers, void();
        int mid = l + r >> 1;
        if ( line[idx[u]]( mid ) > line[id]( mid ) ) std::swap( idx[u], id );
        if ( l == r ) return ;
        if ( line[idx[u]]( l ) > line[id]( l ) ) {
            insert( u << 1, l, mid, id );
        } else if ( line[idx[u]]( r ) > line[id]( r ) ) {
            insert( u << 1 | 1, mid + 1, r, id );
        }
    }

    inline void insert( const Line& l ) {
        line[++lcnt] = l, insert( 1, 1, mxh, lcnt );
    }

    inline LL query( const int u, const int l, const int r, const int x ) {
        if ( ver[u] != vers ) return LINF;
        LL ret = line[idx[u]]( x );
        if ( l == r ) return ret;
        int mid = l + r >> 1;
        if ( x <= mid ) chkmin( ret, query( u << 1, l, mid, x ) );
        else chkmin( ret, query( u << 1 | 1, mid + 1, r, x ) );
        return ret;
    }
} sgt;

#define id( l, r ) ( ( l + r ) | ( l != r ) )
inline void build( const int l, const int r ) {
    int u = id( l, r ); ord[u].resize( r - l + 1 );
    if ( l == r ) return void( ord[u][0] = l );
    int mid = l + r >> 1, lc = id( l, mid ), rc = id( mid + 1, r );
    build( l, mid ), build( mid + 1, r );
    std::merge( ord[lc].begin(), ord[lc].end(), ord[rc].begin(), ord[rc].end(),
      ord[u].begin(), []( const int x, const int y ) { return p[x] < p[y]; } );
}

inline void solve( const int l, const int r ) {
    if ( l == r ) return ;
    int mid = l + r >> 1;
    solve( l, mid ), sgt.clear();
    for ( int u: ord[id( l, r )] ) {
        if ( u <= mid && f[u] != LINF ) {
            sgt.insert( { -2ll * h[u], 1ll * h[u] * h[u] + f[u] } );
        } else if ( u > mid ) {
            chkmin( f[u], 1ll * h[u] * h[u] + a[u]
              + sgt.query( 1, 1, mxh, h[u] ) );
        }
    }
    solve( mid + 1, r );
}
#undef id

int main() {
    n = rint();
    rep ( i, 1, n ) p[i] = rint();
    rep ( i, 1, n ) a[i] = rint();
    rep ( i, 1, n ) h[i] = rint(), mxh = std::max( mxh, h[i] );

    f[1] = a[1];
    rep ( i, 2, n ) f[i] = LINF;
    build( 1, n ), solve( 1, n );
    printf( "%lld
", f[n] );
    return 0;
}