Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28020    Accepted Submission(s): 9864

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

Author

lcy

题目意思：

给n不同种类的物品，每种物品有自己的价值w[i]和个数num[i]，现把全部物品分为两部分，使得两部分总价值最接近，输出两部分总价值。

思路：

物品总价值为sum，那么每部分的总价值最接近sum/2。设一个体积为sum/2的背包，那么问题就转化为选择一些物品使得sum/2的背包中装最大价值的物品，01背包模型。

代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <vector>
 6 #include <queue>
 7 #include <cmath>
 8 #include <set>
 9 using namespace std;
10
11 #define N 55*50*100/2
12
13 int max(int x,int y){return x>y?x:y;}
14 int min(int x,int y){return x<y?x:y;}
15 int abs(int x,int y){return x<0?-x:x;}
16
17 int n;
18 int v[55], num[55];
19 int dp[N];
20
21 main()
22 {
23     int i, j, k;
24     while(scanf("%d",&n)==1&&n>=0){
25         int sum=0;
26         for(i=1;i<=n;i++){
27             scanf("%d %d",&v[i],&num[i]);
28             sum+=v[i]*num[i];
29         }
30         memset(dp,0,sizeof(dp));
31         for(i=1;i<=n;i++){
32             for(k=0;k<=num[i];k++){
33                 for(j=sum/2;j>=k*v[i];j--){
34                     dp[j]=max(dp[j],dp[j-v[i]*k]+v[i]*k);
35                 }
36             }
37         }
38         printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
39     }
40 }