HDU1171：Big Event in HDU(01背包)

最新推荐文章于 2024-08-17 03:35:54 发布

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最新推荐文章于 2024-08-17 03:35:54 发布

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分类专栏： ACM-动态规划文章标签： HDU 动态规划背包

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本文介绍了一个计算机学院与软件学院设施分配的问题，通过01背包算法实现设施价值的均衡分配。输入包括不同种类设施的数量和价值，输出则是两个学院分别获得的设施总价值，确保两者的总价值尽可能接近。

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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：n种设备，每种设备价值为v，每种设备有m个，把这些设备分成两份，每份价值为A、B，尽可能的是A、B相等，并且B总不大于A

思路：01背包问题，B的价值不会超过总价值得1/2，A的价值为总价值减去B的价值

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
#include<stack> 
#include <string>
#include <set>
#include <map>
#include <vector>
using namespace std;
int val[5005],dp[255555];
int main(void)
{
	int n,a,b;
	while(scanf("%d",&n)!=EOF&&n>0)
	{
		int count=0,sum=0;
		memset(val,0,sizeof(val));
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d",&a,&b);
			while(b--)
			{
				val[count++]=a;
				sum+=a; 
			}
		}
		for(int i=0;i<count;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]);      
	}
	return 0;
}