Solution -「AGC 034C」Tests

\(\mathcal{Description}\)

  Link.

  给定非负整数序列 \(\{l_n\},\{r_n\},\{b_n\},X\)，求最小的 \(s\)，使得存在非负整数序列 \(\{a_n\},\{c_n\}\)，满足 \(a_i\le X\)，\(\sum_{i=1}^na_i=s\)，\(c_i\in[l_i,r_i]\)，且

\[\sum_{i=1}^nc_i(a_i-b_i)\ge0
\]

  所有输入均 \(\le10^5\)。

\(\mathcal{Solution}\)

  显然二分 \(s\)，仅需做到检测某个 \(s\) 是否合法。下令 \(w=\sum_{i=1}^nc_ia_i\)。

  假设 \(\{a_n\}\) 已经确定，那么所有满足 \(a_i\ge b_i\) 的 \(c_i=r_i\)，其余 \(c_i=l_i\)。考虑初始时所有 \(a_i=0,c_i=l_i\)，现在把 \(s\) 个 \(1\) 挨个加到一些 \(a_i\) 上。当 \(a_i<b_i\) 时，对 \(w\) 贡献 \(l_i\)（此时 \(c_i\) 仍取 \(l_i\)）；当 \(a_i=b_i\) 时，对 \(w\) 贡献由 \(l_i\) 转为 \(r_i\)（\(c_i\) 变成 \(r_i\)）；继续增加，对 \(w\) 贡献 \(r_i\)。最终仅需比较 \(w\) 和 \(\sum_{i=1}^nl_ib_i\) 的大小。

  所以问题抽象为：有 \(n\) 个分段函数 \(f_{1..n}(x)\)，满足

\[f_i(x)=\begin{cases}l_ix&x\in[0,b_i]\cap\mathbb N\\
l_ib_i+r_i(x-b_i)&x\in(b_i,X]\cap\mathbb N
\end{cases}
\]

  仅需钦定 \(\{x_n\}\)，使得 \(\sum_{i=1}^nf_i(x_i)\) 取最大。

  考虑贪心，不难证明：至多有一个 \(0<x_i<X\)。直接枚举哪一个 \(0<x_i<X\)，贪心地选取最大的 \(f_i(X)\)，即可 \(\mathcal O(n)\) 检测。最终复杂度 \(\mathcal O(n\log\sum_{i=1}^nb_i)\)。

\(\mathcal{Code}\)

/* Clearink */

#include <cstdio>
#include <algorithm>

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

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

typedef long long LL;

const int MAXN = 1e5;
int n, x, b[MAXN + 5], l[MAXN + 5], r[MAXN + 5], ord[MAXN + 5];
LL full[MAXN + 5];

inline LL contr ( const int i, const int s ) {
	return s <= b[i] ? 1ll * s * l[i]
		: 1ll * l[i] * b[i] + 1ll * ( s - b[i] ) * r[i];
}

inline void init () {
	std::sort ( ord + 1, ord + n + 1, []( const int i, const int j ) {
		return contr ( i, x ) > contr ( j, x );
	} );
	rep ( i, 1, n ) full[i] = full[i - 1] + contr ( ord[i], x );
}

inline LL calc ( const LL scr ) {
	/*
	 * let's come up with a greedy algorithm!
	 * */
	int fcnt = scr / x, rest = scr % x;
	LL ret = 0;
	rep ( i, 1, n ) { // score on exam <ord[i]> is <rest>.
		LL cur = contr ( ord[i], rest );
		if ( i > fcnt ) cur += full[fcnt];
		else cur += full[fcnt + 1] - contr ( ord[i], x );
		ret = cur > ret ? cur : ret;
	}
	return ret;
}

int main () {
	n = rint (), x = rint ();
	LL sum = 0, sb = 0;
	rep ( i, 1, n ) {
		ord[i] = i;
		b[i] = rint (), l[i] = rint (), r[i] = rint ();
		sum += 1ll * b[i] * l[i], sb += b[i];
	}
	init ();
	LL lef = 0, rig = sb;
	while ( lef < rig ) {
		LL mid = lef + rig >> 1;
		if ( calc ( mid ) >= sum ) rig = mid;
		else lef = mid + 1;
	}
	printf ( "%lld\n", lef );
	return 0;
}