标签：des style blog color io os java ar strong
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24002 Accepted Submission(s): 8458
Mean:
有n种物品，第一种物品的单价为v1,数量为m1;第二种物品的单价为v2，数量为m2.....现在要你将这些物品分为两堆，使得这两堆物品的价值尽量接近，输出两堆物品的价值。
analyse:
这道题的解法很多：dp,母函数.....，详见《编程之美》。
这里我用了两种方法来做了一下，发现后面的方法比母函数快多了。
Time complexity：O(n^2)
Source code：
母函数代码：
// Memory Time
// 1347K 0MS
// by : Snarl_jsb
// 2014-09-18-18.50
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define N 234567
#define LL long long
using namespace std;
int val[600],cnt[110];
int c1[N],c2[N];
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
// freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);
// freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);
int n;
while(cin>>n&&n>0)
{
long long sum=0;
for(int i=1;i<=n;++i)
{
cin>>val[i]>>cnt[i];
sum+=val[i]*cnt[i];
}
memset(c1,0,sizeof(c1));
memset(c2,0,sizeof(c2));
for(int i=0;i<=cnt[1]*val[1];i+=val[1])
c1[i]=1;
for(int i=2;i<=n;++i)
{
for(int j=0;j<=sum;++j)
{
for(int k=0;k<=cnt[i];++k)
{
c2[val[i]*k+j]+=c1[j];
}
}
for(int j=0;j<=sum;++j)
{
c1[j]=c2[j];
c2[j]=0;
}
}
if(c1[sum/2]==2)
{
cout<<sum/2<<" "<<sum/2<<endl;
continue;
}
int t=sum/2;
int QAQ,TAT;
int minn=987654321;
for(int i=0;i<=sum;++i)
{
if(c1[i])
{
if(abs(sum/2-i)<minn)
{
minn=abs(sum/2-i);
QAQ=i;
}
}
}
TAT=sum-QAQ;
if(QAQ<TAT)
{
QAQ^=TAT^=QAQ^=TAT;
}
cout<<QAQ<<" "<<TAT<<endl;
}
return 0;
}
数学方法：
// Memory Time
// 1347K 0MS
// by : Snarl_jsb
// 2014-09-18-23.05
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define N 1000010
#define LL long long
using namespace std;
int val[N],cnt[N];
int buff[N];
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
// freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);
// freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);
int n;
while(cin>>n&&n>0)
{
long long v,t,idx=0,sum=0;
for(int i=1;i<=n;i++)
{
cin>>v>>t;
val[i]=v,cnt[i]=t;
sum+=val[i]*cnt[i];
while(t--)
{
buff[++idx]=v;
}
}
sort(buff+1,buff+1+idx);
int half=sum/2;
long long ans=0;
for(int i=idx;i>=1;--i)
{
if(ans+buff[i]<=half)
{
ans+=buff[i];
}
}
cout<<sum-ans<<" "<<ans<<endl;
}
return 0;
}
组合数学 - 母函数的变形 --- hdu 1171:Big Event in HDU
标签：des style blog color io os java ar strong
原文地址：http://www.cnblogs.com/acmer-jsb/p/3980377.html
