HDU 1171 Big Event in HDU

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32735    Accepted Submission(s): 11422

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

 

 01背包变形题，有n种物品，每一种价值为value,有m个。求分两堆后，尽可能相等的这两堆价值和分别是多少。

方法:将这所有的物品不管相同不相同，全部分开处理，做背包DP，取sum/2位置的结果就行。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAXN=5005;
const int MAXM=250005;
int weight[MAXN];
int dp[MAXM];
int main()
{
        int n;
        while(scanf("%d",&n)>0&&n>0)
        {
                memset(dp,0,sizeof(dp));
                int len=0;
                int sum=0;
                for(int i=0;i<n;i++)
                {
                        int mWeight,mNum;
                        scanf("%d%d",&mWeight,&mNum);
                        sum+=mNum*mWeight;
                        for(int j=0;j<mNum;j++)
                                weight[len++]=mWeight;
                }
                for(int i=0;i<len;i++)
                        for(int j=sum/2;j>=weight[i];j--)
                                dp[j]=max(dp[j],dp[j-weight[i]]+weight[i]);
                printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
        }
        return 0;
}