Big Event in HDU(HDU-1171)

最新推荐文章于 2021-09-08 21:36:57 发布

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最新推荐文章于 2021-09-08 21:36:57 发布

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解决HDU计算机学院和软件学院的物资平均分配问题，通过动态规划算法实现物资价值的最接近划分。

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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): 38716 Accepted Submission(s): 13395

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

题目大意：

HDU要分开计算机学院和软件学院，需要对现有物资进行尽可能地平均分配，并按照从小到大的顺序输出出来。数据以负数结束。

真是呵呵了，还以为是以-1结束，超时了好几次。。。。。模板题。

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[5353],dp[255555];
int amount = 0,sum,n;
void solve()
{
	for(int i=0 ; i<amount ; i++)
		for(int k=sum/2 ; k>=a[i] ; k--)
			dp[k] = max(dp[k],dp[k-a[i]]+a[i]);
}
int main()
{
	while(~scanf("%d",&n),n>0)
	{
		sum=0;
		amount = 0;
		for(int i=0 ; i<n ; i++)
		{
			int p,q;
			scanf("%d%d",&p,&q);
			sum += p*q;
			while(q--)
			a[amount++] = p;
		}
		memset(dp,0,sizeof(dp));
		solve();
		printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
	}
	return 0;
}

题目传送门。。。。。