HDU1171(01背包)

01背包问题解析与代码实现

最新推荐文章于 2020-04-02 16:55:44 发布

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本文深入探讨了01背包问题的经典算法，通过一个具体的编程实例，详细讲解了如何利用动态规划解决设施分配问题，确保计算机学院和软件学院在设施分割中获得尽可能平等的价值。

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

#include<iostream>
#include <algorithm>
using namespace std;
int val[5005];

int dp[255555];

int main()

{

    int n,a,b,l,sum;

    while（cin>>n,n>0)

    {

        memset(val,0,sizeof(val));

        memset(dp,0,sizeof(dp));

        l = 0;

        sum = 0;

        for(int i = 0;i<n;i++)

        {

            scanf("%d%d",&a,&b);

            while(b--)

            {

                val[l++] = a;//将价值存入数组

                sum+=a;

            }

        }

        for(int i = 0;i<l;i++)

        {

            for(int j = sum/2;j>=val[i];j--)//01背包

            {

                dp[j] = max(dp[j],dp[j-val[i]]+val[i]);

            }

        }

        printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);

    }

 

    return 0;

}

30 1
-1

Sample Output

20 10
40 40
问题分析：01背包

AC代码：