\(\mathcal{Description}\)
Link.
把 \(n\) 種零食分給 \(m\) 個人,第 \(i\) 種零食有 \(a_i\) 個;第 \(i\) 個人得到同種零食數量不超過 \(b_i\),總數量不超過 \(c_i\),求最多分出的零食數量。
\(n,m\le2\times10^5\)。
\(\mathcal{Solution}\)
很容易看出這是網絡流模型:
- 源點 \(S\) 連向每種零食 \(i\),容量 \(a_i\);
- 零食 \(i\) 連向人 \(j\),容量 \(b_j\);
- 人 \(j\) 連向彙點 \(T\),容量 \(c_j\)。
\(\mathcal{Code}\)
/*~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 )
typedef long long LL;
inline void chkmin( LL& a, const LL b ) { b < a && ( a = b ); }
const int MAXN = 2e5;
int n, m, b[MAXN + 5], ord[MAXN + 5];
LL a[MAXN + 5], c[MAXN + 5];
int main() {
scanf( "%d %d", &n, &m );
rep ( i, 1, n ) scanf( "%lld", &a[i] );
rep ( i, 1, m ) scanf( "%d", &b[i] ), ord[i] = i;
rep ( i, 1, m ) scanf( "%lld", &c[i] );
std::sort( a + 1, a + n + 1,
[]( const LL u, const LL v ) { return u > v; } );
std::sort( ord + 1, ord + m + 1, []( const int u, const int v )
{ return 1ull * c[u] * b[v] < 1ull * c[v] * b[u]; } );
LL sa = 0, sb = 0, sc = 0, ans = 1ll << 60;
rep ( i, 1, n ) sa += a[i];
rep ( i, 1, m ) sb += b[i];
for ( int i = 0, j = 1; i <= n; ++i ) {
sa -= a[i];
for ( ; j <= m && 1ll * i * b[ord[j]] > c[ord[j]];
sb -= b[ord[j]], sc += c[ord[j++]] );
chkmin( ans, sa + i * sb + sc );
}
printf( "%lld\n", ans );
return 0;
}