Big Event in HDU （HDU_1171) 01背包

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32236    Accepted Submission(s): 11276

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

210 120 1310 1 20 230 1-1

 

Sample Output

20 1040 40

 

Author

lcy

题目大意：有n种物品，给出这n件物品的价值和件数，求如何将这些物品按价值尽可能平分。如果不能平分，前者分得较多。

解题思路：按道理应该用完全背包来做，不过用01背包也能AC（可能是后台的样例太弱）。01背包，物品的体积和价值都定位W，背包的容量和为sum/2，然后将所有物品有重复地存入数组a中，其中动态转移方程为：dp[j]=max(dp[j],dp[j-a[i]]+a[i]);

#include"iostream"#include"cstdio"#define MAX_N 50*100*50+10  //最大体积 N*M*V#define MAX_C 50*100+10//最大物品树 N*M using namespace std;int dp[MAX_N/2];int a[MAX_C];int max(int x,int y){return x>y?x:y;}int main(){int n;while(scanf("%d",&n)!=EOF){if(n<=0) break;int t,k;int num=0,sum=0;for(int i=1,j=1;i<=n;i++){scanf("%d%d",&t,&k);sum+=t*k; num+=k;for(;k>0;k--){a[j]=t;j++;}}for(int i=0;i<=sum/2;i++){dp[i]=0;}for(int i=1;i<=num;i++){for(int j=sum/2;j>=a[i];j--){dp[j]=max(dp[j],dp[j-a[i]]+a[i]);}}printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);}return 0;}