Big Event in HDU -背包问题

喜欢ヅ旅行 2022-06-06 09:06 86阅读 0赞

# 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 ##

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

## Sample Output ##

20 10
    40 40

题意：有N种物品，每种物品的价值和重量分别是V,M。平分这些物品。

把此多重背包问题转化为01背包问题。背包限制最多只能装sum/2的量。虽然时间长，但是也可以通过的。

#include <bits/stdc++.h>
    using namespace std;
    int N,Q;
    int M;
    int V;//背包最大容量
    int value[5005];//物品价值
    int f[255555];
    int ZeroOnePack()
    {
        memset(f,0,sizeof(f));
        //递推
        for (int i = 0;i < Q;i++) //枚举物品(注意物品的编号)
        {
            for (int j = V;j >= value[i];j--)
            {
                f[j] = max(f[j],f[j - value[i]] + value[i]);
                //cout << j << " " << f[j] << endl;
            }
        }
        return f[V];
    }
    int main()
    {
        int v,sum;
        while(~scanf("%d",&N))
        {
            if(N<0)break;
            Q=0;
            sum=0;
            for(int i=0;i<N;i++)
            {
                scanf("%d%d",&v,&M);
                while(M--)
                {
                    value[Q++]=v;
                    sum=sum+v;
                }
                V=sum/2;
            }
            printf("%d %d\n",sum-ZeroOnePack(),ZeroOnePack());
        }
        return 0;
    }