hdoj 2844 Coins【多重背包】【dp】

最新推荐文章于 2019-07-08 15:32:10 发布

坚定的水手

最新推荐文章于 2019-07-08 15:32:10 发布

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本文链接：https://blog.youkuaiyun.com/liu6886/article/details/49700673

本文介绍了一种解决多重背包问题的方法，通过完全背包和二进制拆分进行01背包来处理不同面值硬币组合的可能性，以计算出能够支付的价格总数。

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Coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10300 Accepted Submission(s): 4106

Problem Description

Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.

You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

Input

The input contains several test cases. The first line of each test case contains two integers n(1 ≤ n ≤ 100),m(m ≤ 100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1 ≤ Ai ≤ 100000,1 ≤ Ci ≤ 1000). The last test case is followed by two zeros.

Output

For each test case output the answer on a single line.

Sample Input

  
   3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output

8
4

Source

2009 Multi-University Training Contest 3 - Host by WHU

题意：

你有n种面值的硬币，告诉你每种硬币的数量，问这些硬币能拼成多少不超过m的面额。

思路：

多重背包。当某种硬币的总额大于背包容量时进行完全背包，否则进行二进制拆分进行01背包。最后统计dp[i]==i的数量，即恰好能凑出金额为i的情况总数sum。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[100500];
int val[500],num[500];
void CompletePack(int v,int w, int m)
{
	for(int i = v; i <= m; i++)
		dp[i] = max(dp[i], dp[i-v]+w);
} 
void ZeroOnePack(int v,int w,int m)
{
	for(int i = m; i >= v; i--)
	{
		dp[i] = max(dp[i], dp[i-v]+w);
	}
}
void MultiPack(int v, int w, int m, int c)
{
	if(v*c > m)	CompletePack(v,w,m);
	else
	{
		int k = 1;
		while(k < c)
		{
			ZeroOnePack(k*v, k*w, m);
			c -= k;
			k *= 2;
		}
		ZeroOnePack(c*v, c*w, m);
	}	
	
}
int main()
{
	int n,m;
	while(~scanf("%d%d",&n,&m),n+m)
	{
		int i,j;
		memset(dp, 0, sizeof(dp));
		for(i = 1; i <= n; i++)
			scanf("%d", &val[i]);
		for(i = 1; i <= n; i++)
			scanf("%d", &num[i]);
		for(i = 1; i <= n; i++)
			MultiPack(val[i],val[i],m,num[i]);
		int sum = 0;
		for(i = 1; i <= m; i++)
			if(dp[i]==i)	sum++;
		printf("%d\n", sum);			
	}
	return 0;
}