HDU 1171 Big Event in HDU(多重背包)

最新推荐文章于 2020-04-11 15:07:56 发布

KinFungYu

最新推荐文章于 2020-04-11 15:07:56 发布

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分类专栏： hdu 动态规划背包问题文章标签：动态规划多重背包问题

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本文探讨了多重背包问题的解决策略，并通过一个具体的案例分析了如何在HDU计算机学院与软件学院设施分配问题中应用该算法。通过输入不同类型的设施数量、价值和对应数量，输出了两个学院将获得的价值大致相等的分配方案。

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Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15812 Accepted Submission(s): 5574

Problem Description

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

Output

For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

Sample Input

Sample Output

20 10
40 40
 
多重背包问题
AC代码：
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

int dp[250000+5];

int MAX(int a,int b)
{
    return a>b?a:b;
}
int MIN(int a,int b)
{
    return a<b?a:b;
}
int main()
{
  int w[55],a[55];
  int t;
  while(cin>>t)
  {
      if(t<0)
      break;
      int sum=0;
      for(int i=1;i<=t;i++)
      {
          cin>>w[i]>>a[i];
          sum+=w[i]*a[i];
      }
      int sum1=sum/2;
      memset(dp,0,sizeof(dp));
      for(int i=1;i<=t;i++)
      for(int j=1;j<=a[i];j++)
      for(int p=sum1;p>=w[i];p--)
      {
          dp[p]=MAX(dp[p],dp[p-w[i]]+w[i]);
      }
      cout<<MAX(dp[sum1],sum-dp[sum1])<<" "<<MIN(dp[sum1],sum-dp[sum1])<<endl;
  }

return 0;
}