HDU1171--Big Event in HDU（多重背包）

Posted 2020-08-17 Pic

 

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

这是一个多重背包的题目（将物品的原价值即为放入背包中物品的价值与花费，因为当不超过背包容量且价值之和最大时，即为不超过背包容量且花费之和最大（花费最接近背包容量），即为这些物品的原价值之和最接近背包容量。此处将物品总价值的一半看为背包容量即可。

这里有一个我见过的最诡异的八阿哥。。。汗。。。

这道题是以负数作为输入的结束标志的，而不只局限于-1， 被惯性思维坑了。。。。

代码：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 const int MAXN=250010;
 8 const int inf=99999999;
 9 int v[MAXN];
10 int dp[MAXN];
11 int num[55];
12 int main()
13 {
14     //freopen("data.in","r",stdin);
15     std::ios::sync_with_stdio(false);
16     std::cin.tie(0);
17     int n;
18     int sum;
19     int cc;
20     while(cin>>n&&n>=0){
21         sum=0;
22         memset(dp,0,sizeof(dp));
23         for(int i=0;i<n;i++){
24             cin>>v[i]>>num[i];
25             sum+=(v[i]*num[i]);
26         }
27         cc=sum/2;
28         for(int i=0;i<n;i++){
29             for(int k=0;k<num[i];k++){
30                 for(int j=cc;j>=v[i];j--){
31                         dp[j]=max(dp[j],dp[j-v[i]]+v[i]);
32                 }
33             }
34         }
35         cout<<sum-dp[cc]<<" "<<dp[cc]<<endl;
36     }
37 }