Big Event in HDU（HDU-1171）

最新推荐文章于 2020-09-04 01:51:25 发布

Alex_McAvoy

最新推荐文章于 2020-09-04 01:51:25 发布

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分类专栏： # 动态规划——背包问题 # HDU

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本文介绍了一个关于0-1背包问题的实际案例，通过解决计算机学院与软件学院设施分配问题来展示如何运用0-1背包算法。具体实现中，首先统计所有设施的总价值，并将其平分为两部分，然后采用0-1背包算法找到最接近平分值的方案。

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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

2
10 1
20 1
3
10 1
20 2
30 1
-1

Sample Output

20 10
40 40

思路：统计总和，将总和平分后，就是0-1背包问题了

Source Program

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int dp[250001];
int main()
{
    int n;
    int v[51],num[51];
    int sum;
    int i,j,k;

    while(cin>>n && n>0)
    {
        /*初始化*/
        sum=0;
        memset(dp,0,sizeof(dp));

        for(i=1;i<=n;i++)
        {
            cin>>v[i]>>num[i];//输入价值与数量
            sum+=v[i]*num[i];//统计总价值

        }

        for(i=1;i<=n;i++)//物品价值
            for(j=1;j<=num[i];j++)//每个物品数量
                for(k=sum/2;k>=v[i];k-=v[i])//把价值看作体积，从中间开始
                    dp[k]=max(dp[k],dp[k-v[i]]+v[i]);
        cout<<sum-dp[sum/2]<<" "<<dp[sum/2]<<endl;
    }
    return 0;
}