题目描述
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
分析
根据分析,本题是n件物品,对应价值v和个数m,求一半总价值的最大值,是多重背包问题。
可以使用二进制优化多重背包加快运算速度。
C++ 代码
#include<bits/stdc++.h>
using namespace std;
const int N = 1e4+10;
int f[N*10],w[N],n,a,c;
int main()
{
ios::sync_with_stdio(false);
while(cin>>n)
{
if(n<=0) break;
int sum=0,co=0;
memset(f,0,sizeof f);
for(int i=0;i<n;i++)
{
cin>>a>>c;
sum+=(c*a);
for(int k=1;k<=c;k*=2) //每个物品划分为二进制形式
{
w[co++]=a*k;
c-=k;
}
if(c) w[co++]=a*c;
}
int v=sum/2; //要求一半的体积下的最大值
for(int i=0;i<co;i++) //二进制优化后就变成了0-1背包问题
{
for(int j=v;j>=w[i];j--)
f[j]=max(f[j],f[j-w[i]]+w[i]);
}
cout<<max(sum-f[v],f[v])<<" "<<min(sum-f[v],f[v])<<endl; //输出此时的最大值和最小值
}
return 0;
}