1. 程式人生 > >[BZOJ1531] [POI2005] Bank notes [多重揹包]

[BZOJ1531] [POI2005] Bank notes [多重揹包]

[ L i n k \frak{Link} ]


把樸素多重揹包的式子稍作轉化,得到:
f

j = m a x { f k
k v a l u e
      k [ j c o u n t i , j ] } + j v a l u e ,    j c o s t i + R e m a i n d e r = j \mathcal{f_{j}=max\{f_{k'}-k'value\;|\;k'\in[j'-count_i,j']\}+j'value,\;j'cost_i+Remainder=j} (列舉 R e m a i n d e r \mathcal{Remainder}
看了半天感覺不太對結果發現這道題也是代價等於價值,,


#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<cstring>
using namespace std;
int n, m;
int a[205];
int c[205];
pair<int, int> q[20005];
int qh, qt;
int f[20005];
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
	}
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &c[i]);
	}
	scanf("%d", &m);
	memset(f, 0x3f, sizeof(f));
	f[0] = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j < a[i]; ++j) {
			qh = 1;
			qt = 0;
			for (int x, k = 0; ; ++k) {
				if ((x = k * a[i] + j) > m) break;
				while (qh <= qt && q[qh].second < k - c[i]) {
					++qh;
				}
				while (qh <= qt && q[qt].first >= f[x] - k) {
					--qt;
				}
				q[++qt] = make_pair(f[x] - k, k);
				f[x] = min(f[x], q[qh].first + k);
			}
		}
	}
	printf("%d", f[m]);
	return 0;
}