(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_ile X),(sum_{i=1}^na_i=s),(c_iin[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_ige 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)=egin{cases}l_ix&xin[0,b_i]capmathbb N\
l_ib_i+r_i(x-b_i)&xin(b_i,X]capmathbb 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(nlogsum_{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
", lef );
return 0;
}