hdu 1171 Big Event in HDU 多重背包

最新推荐文章于 2021-02-01 08:04:36 发布

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文章回顾了计算机学院在2002年被分为计算机学院和软件学院的重要历史事件，详细介绍了设施分配过程和计算出两个学院各自获得的价值。

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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

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
struct node
{
    int value,num;
} q[55];
int f[250000+5];
int main()
{
    int n;
    int total;
    while(~scanf("%d",&n))
    {
        if(n<0)
            break;
        total=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d %d",&q[i].value,&q[i].num);
            total+=q[i].value*q[i].num;
        }
        int total1=total/2;
        memset(f,0,sizeof(f));
        for(int i=1; i<=n; i++)
            for(int k=1; k<=q[i].num; k++)
                for(int p=total; p>=q[i].value; p--)
                    f[p]=max(f[p],f[p-q[i].value]+q[i].value);
        printf("%d %d\n",max(f[total1],total-f[total1]),min(f[total1],total-f[total1]));
    }
    return 0;
}