C - Big Event in HDU HDU - 1171（（多重背包+二进制优化））

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01背包问题同时被 3 个专栏收录

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多重背包问题

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博客讲述了HDU计算机学院在2002年拆分为计算机学院和软件学院，设施需平分。介绍了设施评估规则，不同价值为不同种类。给出输入输出要求，输入含多测试用例，以负数开头终止，输出两学院应得设施价值，且尽量相等，A不小于B。

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

#include<stdio.h>
#include<string.h>

int f[250005],a[10005];

int max(int x,int y)
{
    return x>y?x:y;
}

int main()
{
    int n,V,sum,c,x,y,i,j;
    while(scanf("%d",&n)&&n>=0)
    {
        memset(f,0,sizeof(f));
        memset(a,0,sizeof(a));
        sum=0;
        c=0;
        for(i=1; i<=n; i++)
        {
            scanf("%d%d",&x,&y);
            if(y>1)
            {
                int t=1;
                while(y>=t)
                {
                    a[++c]=x*t;
                    sum+=x*t;
                    y=y-t;
                    t=t*2;
                }
                a[++c]=x*y;
                sum+=x*y;
            }
            else
            {
                a[++c]=x*y;
                sum+=x*y;
            }

        }

        V=sum/2;
        for(i=1; i<=c; i++)
        {
            for(j=V; j>=a[i]; j--)
            {
                f[j]=max(f[j],f[j-a[i]]+a[i]);
            }
        }
        printf("%d %d\n",sum-f[V],f[V]);
    }
    return 0;
}