HDU1171--Big Event in HDU（多重背包）

 

Big Event in HDU

 

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1139 Accepted Submission(s): 444

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

這是一個多重背包的題目（將物品的原價值即為放入背包中物品的價值與花費，因為當不超過背包容量且價值之和最大時，即為不超過背包容量且花費之和最大（花費最接近背包容量），即為這些物品的原價值之和最接近背包容量。此處將物品總價值的一半看為背包容量即可。

這里有一個我見過的最詭異的八阿哥。。。汗。。。

這道題是以負數作為輸入的結束標志的，而不只局限於-1， 被慣性思維坑了。。。。

代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int MAXN=250010;
const int inf=99999999;
int v[MAXN];
int dp[MAXN];
int num[55];
int main()
{
    //freopen("data.in","r",stdin);
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int n;
    int sum;
    int cc;
    while(cin>>n&&n>=0){
        sum=0;
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++){
            cin>>v[i]>>num[i];
            sum+=(v[i]*num[i]);
        }
        cc=sum/2;
        for(int i=0;i<n;i++){
            for(int k=0;k<num[i];k++){
                for(int j=cc;j>=v[i];j--){
                        dp[j]=max(dp[j],dp[j-v[i]]+v[i]);
                }
            }
        }
        cout<<sum-dp[cc]<<" "<<dp[cc]<<endl;
    }
}