HDU2191 多重背包

题意：

中文不解释

要点：

多重背包裸题，主要是用了两种做法，一种直接按个数全拆成01背包，复杂度是O(V*Σn[i])；一种用二进制拆成复杂度为O(V*Σlog n[i])。

复杂度是O(V*Σn[i])：

16571118

2016-03-16 15:17:11

Accepted

2191

31MS

1416K

512 B

G++

seasonal

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define max(a,b) a>b?a:b
int v[150], w[150], c[150],dp[150];
int n, m;

int main()
{
	int t,i,j,k;
	scanf("%d", &t);
	while (t--)
	{
		memset(dp, 0, sizeof(dp));
		scanf("%d%d", &m, &n);
		for (int i = 0; i < n; i++)
			scanf("%d%d%d", &w[i], &v[i], &c[i]);
		for (i = 0; i < n; i++)
			for (j = 0; j < c[i]; j++)//拆成c[i]个01背包
				for (k = m; k >= w[i]; k--)
					dp[k] = max(dp[k], dp[k - w[i]] + v[i]);
		printf("%d\n", dp[m]);
	}
	return 0;
}

复杂度为O(V*Σlog n[i])：

16570781

2016-03-16 14:37:38

Accepted

2191

0MS

1432K

985 B

G++

seasonal

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define max(a,b) a>b?a:b
int v[1500], w[1500], c[1500],dp[1500];
int n, m;

void one_pack(int cost, int weight)//01背包
{
		for (int j = m; j >= cost; j--)
			dp[j] = max(dp[j], dp[j - cost] + weight);
}
void complete_pack(int cost, int weight)//完全背包
{
	for (int j = cost; j <= m; j++)
		dp[j] = max(dp[j], dp[j -cost] + weight);
}
void multiple_pack(int cost, int weight, int count)
{
	int i;
	if (cost*count >= m)//如果一开始就count*cost>=m，可以看成无限种也就是完全背包
	{
		complete_pack(cost, weight);
	}
	else
	{
		int k=1;
		while (k < count)
		{
			one_pack(k*cost, k*weight);//相当于01背包
			count -= k;
			k *= 2;				//利用二进制思想可以表示1-count的所有值
		}	
		one_pack(count*cost, count*weight);//最后处理剩下的个数
	}
}


int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		memset(dp, 0, sizeof(dp));
		scanf("%d%d", &m, &n);
		for (int i = 0; i < n; i++)
			scanf("%d%d%d", &w[i], &v[i], &c[i]);
		for (int i = 0; i < n; i++)
			multiple_pack(w[i], v[i], c[i]);
		printf("%d\n", dp[m]);
	}
	return 0;
}