HDU Big Event in HDU

最新推荐文章于 2020-09-04 01:51:25 发布

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本文详细阐述了如何通过算法解决背包问题，并提供了一个具体的C++代码实例，该实例涉及了计算机学院和软件学院设施分配的场景。

Big Event in HDU

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

这题是一个典型的背包问题，这里的背包最大容积为w[i](1<=i<=n)的一半。而背包所能取的最大物品数为

n[i](1<=i<=n)的一半。

#include<iostream>
using namespace std;
int main()
{
static bool b[2500001];
static int q[2500001];
int n,p[55],w[55],sum,i,j,k,max,min,num,s;
while(cin>>n&&n>=0)
{
  sum=0;
  num=0;
  for(i=1;i<=n;i++)
  {
   cin>>p[i]>>w[i];
   sum+=p[i]*w[i];
   num+=w[i];
  }
  s=sum;
  sum=sum/2;
  num=num/2;
  memset(b,false,sizeof(b));
  b[0]=true;
  q[0]=0;
  for(i=1;i<=n;i++)
   for(k=1;k<=w[i];k++)
    for(j=sum;j>=0;j--)
    {
     if(b[j]&&j+p[i]<=sum&&q[j]<num)
     {
      b[j+p[i]]=true;
      q[j+p[i]]=q[j]+1;
     }
    }

  for(i=sum;i>=0;i--)
   if(b[i]==true)
    break;
        if(s-i>i)
  {
   max=s-i;
   min=i;
  }
  else
  {
   max=i;
   min=s-i;
  }
  printf("%d %d/n",max,min);

}
return 0;
}