动态规划：HDU2844-Coins（多重背包的二进制优化）

Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory
Limit: 32768/32768 K (Java/Others)

Total Submission(s): 9826    Accepted Submission(s): 3916

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 

题意：有n中面值的硬币每种面值的硬币数量已知问这些硬币能组成1-m中的几种多重背包已m作为背包容量在1-m中当有物品的价值等
于背包容量时即能组成该类型。

解题心得：

1、就是一个多重背包的二进制优化问题，题目要求的是一个最后表价格可能的个数。

2、关于多重背包的二进制优化问题（多重背包的二进制优化）

#include<bits/stdc++.h>
using namespace std;
const int maxn1 = 1e5+100;
const int maxn2 = 110;
int n,m;
bool dp[maxn1];
struct Coin
{
    int num;
    int va;
} c[maxn2];

int pre_coin()
{
    int sum = 0;
    for(int i=1; i<=n; i++)
        scanf("%d",&c[i].va);
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&c[i].num);
        sum += c[i].num * c[i].va;
    }
}
int main()
{
    while(scanf("%d%d",&n,&m) && n+m)
    {
        memset(dp,0,sizeof(dp));
        dp[0] = true;
        pre_coin();
        for(int i=1; i<=n; i++)
        {
            int cnt = 0;
            for(int k=1; k<=c[i].num; k*=2)//k*=2二进制优化
            {
                cnt = k*c[i].va;
                for(int j=m; j>=cnt; j--)
                    if(dp[j-cnt])
                        dp[j] = true;
                c[i].num -= k;
            }

            if(cnt)
            {
                cnt = c[i].num * c[i].va;
                for(int j=m; j>=cnt; j--)
                {
                    if(dp[j-cnt])
                        dp[j] = true;
                }
            }
        }

        int Sum = 0;//记录可能是手表价格的个数
        for(int i=m; i>=1; i--)
            if(dp[i])
                Sum ++;
        printf("%d\n",Sum);
    }
}