HDU 1171 01背包变形

最新推荐文章于 2019-07-25 17:01:08 发布

原创最新推荐文章于 2019-07-25 17:01:08 发布 · 223 阅读

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#01背包变形 #dp

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本文解析了一道关于计算机学院和软件学院设施分配的问题。通过将不同数量的设施转换为01背包问题进行求解，确保两个学院分配到的设施价值尽可能接近且前者不小于后者。

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Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43941 Accepted Submission(s): 15113

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

Sample Output

20 10
40 40


一开始看这道题，既不是完全背包也不是01背包，以为要用到多重背包，看了看大佬的博客，才发现可以变成01背包来解决，只要把机器的价值分开来算就行了，就是时间不算短，等会了新的方法在更新吧。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int val[5050],dp[250050];
int main()
{
	int n;
	while(scanf("%d",&n) && n>0)
	{
		memset(dp,0,sizeof(dp));
		int a,b,num=0,sum=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d",&a,&b);
			sum+=a*b; 
			while(b--)
				val[num++]=a; 
		} 
		for(int i=0;i<num;i++)
		{
			for(int j=sum/2;j>=val[i];j--)
				dp[j]=max(dp[j],dp[j-val[i]]+val[i]);
		}
		printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
	}
}