poj 1742(多重背包)

最新推荐文章于 2020-02-27 09:15:19 发布

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本文介绍了一种解决多重背包问题的高效算法，并提供了一个具体的C++实现案例。该算法通过优化选择硬币支付方式，计算在给定条件下可以支付的不同价格数目。

Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 33224		Accepted: 11279

Description

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box 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

LouTiancheng@POJ

裸的多重背包不能用自带的max函数时间耗费太大了会超时

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

int a[200];
int num[200];
int dp[100005];
int n,m;
void ZeroOnePack(int w,int v)
{
    for(int i=m;i>=w;i--)
        if(dp[i-w]+v>dp[i])
            dp[i]=dp[i-w]+v;
}

void CompletePack(int w,int v)
{
    for(int i=w;i<=m;i++)
        if(dp[i-w]+v>dp[i])
            dp[i]=dp[i-w]+v;
}

void MultiplePack()
{
    for(int i=1;i<=n;i++)
    {
        if(a[i]*num[i]>=m)
            CompletePack(a[i],a[i]);
        else
        {
            int k=1;
            while(k<num[i])
            {
                ZeroOnePack(a[i]*k,a[i]*k);
                num[i]-=k;
                k*=2;
            }
            ZeroOnePack(a[i]*num[i],a[i]*num[i]);
        }
    }
}

int main()
{
    while(~scanf("%d%d",&n,&m)&&(n+m))
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        for(int i=1;i<=n;i++)
            scanf("%d",num+i);
        MultiplePack();
        int res=0;
        for(int i=1;i<=m;i++)
            if(dp[i]==i)
                res++;
        printf("%d\n",res);
    }
    return 0;
}