hdu 1171 多重背包

题意：给出价值和数量，求能分开的最近的两个总价值，例如10,20*2,30，分开就是40，40

链接：点我

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
using namespace std;
#define MOD 1000000007
const int INF=0x3f3f3f3f;
const double eps=1e-5;
typedef long long ll;
#define cl(a) memset(a,0,sizeof(a))
#define ts printf("*****
");
const int MAXN=250010;
int n,m,tt;
int a[MAXN],b[MAXN];
int dp[MAXN];

int nValue;
//0-1背包，代价为cost,获得的价值为weight
void ZeroOnePack(int cost,int weight)
{
    for(int i=nValue;i>=cost;i--)
        dp[i]=max(dp[i],dp[i-cost]+weight);
}
//完全背包，代价为cost,获得的价值为weight
void CompletePack(int cost,int weight)
{
    for(int i=cost;i<=nValue;i++)
        dp[i]=max(dp[i],dp[i-cost]+weight);
}
//多重背包
void MultiplePack(int cost,int weight,int amount)
{
    if(cost*amount>=nValue)CompletePack(cost,weight);
    else
    {
        int k=1;
        while(k<amount)
        {
            ZeroOnePack(k*cost,k*weight);
            amount-=k;
            k<<=1;
        }
        ZeroOnePack(amount*cost,amount*weight);
    }
}
int main()
{
    int i,j,k;
    #ifndef ONLINE_JUDGE
    freopen("1.in","r",stdin);
    #endif
    while(scanf("%d",&n)!=EOF)
    {
        if(n<0) break;
        nValue=0;
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&a[i],&b[i]);
            nValue+=a[i]*b[i];
        }
        cl(dp);
        for(i=0;i<n;i++)
        {
            MultiplePack(a[i],a[i],b[i]);
        }
        int v=nValue/2;
        while(dp[v]!=v) v--;
        printf("%d %d
",nValue-v,v);
    }
}