HDU 1171 Big Event in HDU （动态规划、01背包）

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57986    Accepted Submission(s): 19484

 

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

题目大意与分析

给出一堆东西的重量和个数，要尽可能的将物品分成重量接近的两堆A和B,且A的重量不小于B

求以sum/2为容量的背包即可，求出来的值就是B sum减去B就是A

代码

#include <bits/stdc++.h>
using namespace std;
 
int val[5005];
int dp[255555];
 
int main()
{
    int n,i,j,a,b,l,sum;
    while(~scanf("%d",&n),n>0)
    {
        memset(val,0,sizeof(val));
        memset(dp,0,sizeof(dp));
        l = 0;
        sum = 0;
        for(i = 0;i<n;i++)
        {
            scanf("%d%d",&a,&b);
            while(b--)
            {
                val[l++] = a;
                sum+=a;
            }
        }
        for(i = 0;i<l;i++)
        {
            for(j = sum/2;j>=val[i];j--)
            {
                dp[j] = max(dp[j],dp[j-val[i]]+val[i]);
            }
        }
        printf("%d %d
",sum-dp[sum/2],dp[sum/2]);
    }
 
    return 0;
}