HDU-1171 Big Event in HDU（01背包）

最新推荐文章于 2022-01-13 08:42:39 发布

原创最新推荐文章于 2022-01-13 08:42:39 发布 · 152 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++

动态规划同时被 2 个专栏收录

85 篇文章

订阅专栏

背包

38 篇文章

订阅专栏

本文介绍了一个关于 HDU 计算机学院历史变迁的问题，涉及如何公平地分配设施资源，通过背包算法模拟决策过程。核心在于理解01背包问题的应用，目标是使两所学院得到尽可能接近的价值分配。

Big Event in HDU
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 64528 Accepted Submission(s): 21654

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
Author

lcy

Recommend

这个题的思考点主要还是在能想到，背包的容量是（总的钱 / 2 ）

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
const int N=5000010;
int f[N];
int a,b,m;
int sum,ans,cou;
int v[5010];

int main()
{
	int n;
	while(cin>>n,n>=0)
	{
		memset(f,0,sizeof(f));
		sum=0;//需要的分的钱的总数 
		cou=0;
		for(int i=1;i<=n;i++)
		{
			cin>>a>>b;
			while(b--)
			{
				v[++cou]=a;
				sum+=a;
			}
		}
		
		for(int i=1;i<=cou;i++)//钱的总数目
		{
			for(int j=sum/2;j>=v[i];j--)//01背包 
			{
				f[j]=max(f[j],f[j-v[i]]+v[i]);
			}
		}
		int res1=f[sum/2];
		int res2=sum-res1;
		if(res1>res2)
		cout<<res1<<" "<<res2<<endl;
		else cout<<res2<<" "<<res1<<endl;
	}
	return 0;
}