Big Event in HDU
提供多重背包的一些思路
Description
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
A test case starting with a negative integer terminates input and this test case is not to be processed.
Output
Sample Input
Sample Output
网上的代码：
#include
#include
#include
using namespace std;
int dp[100000],sum,ans;
struct st
{
int v;
int m;
}data[100000];
void full(int x)
{
for(int i=data[x].v;i<=ans;i++)
dp[i]=max(dp[i],dp[i-data[x].v]+data[x].v);
}
void one(int x)
{
for(int j=1;j<=data[x].m;j++)
for(int i=ans;i>=data[x].v;i--)
dp[i]=max(dp[i],dp[i-data[x].v]+data[x].v);
}
int main()
{
int i,j,n;
while(scanf("%d",&n)&&(n>0))
{
memset(dp,0,sizeof(dp));
sum=0;
for(i=1;i<=n;i++)
{
scanf("%d%d",&data[i].v,&data[i].m);
sum+=data[i].v*data[i].m;
}
ans=sum/2;
for(i=1;i<=n;i++)
{
if(data[i].v*data[i].m>=ans)
full(i);
else
one(i);
}
printf("%d %d\n",sum-dp[ans],dp[ans]);
}
return 0;
}
//母函数方法：
/*注意将数组a,s清零，WA了好几次，测试数据都过。。无语。
*/
#include
#include
int a[250010],s[250010];
int v[55],m[55];
int main()
{
int n,i,j,k,sum,ans;
while(scanf("%d",&n)&&n>0)
{
sum=0;
memset(s,0,sizeof(s));
memset(a,0,sizeof(a));
for(i=1;i<=n;i++)
{
scanf("%d%d",&v[i],&m[i]);
sum+=v[i]*m[i];
}
for(i=0;i<=v[1]*m[1];i+=v[1])//注意变化。
{
s[i]=1;
}
for(i=2;i<=n;i++)
{
for(j=0;j<=sum;j++)
{
for(k=0;k+j<=sum&&k<=v[i]*m[i];k+=v[i])
{
a[k+j]+=s[j];
}
}
for(k=0;k<=sum;k++)
{
s[k]=a[k];
a[k]=0;
}
}
for(i=sum/2;i>=0;i--)
{
if(s[i])
{
printf("%d %d\n",sum-i,i);
break;
}
}
}
return 0;
}
我的代码：
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<climits>
#include<queue>
#include<vector>
#include<map>
#include<sstream>
#include<set>
#include<stack>
using namespace std;
//typedef long long ll;
//ifstream fin("input.txt");
//ofstream fout("output.txt");
//fin.close();
//fout.close();
int val[60];
int num[60];
int n;
int dp[270000];
int main()
{
int i,j,k;
while(scanf("%d",&n)==1&&n>0)
{
int cnt=0;
for(i=1;i<=n;i++)
{
scanf("%d%d",&val[i],&num[i]);
cnt+=num[i]*val[i];
}
int sum=cnt/2;
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
{
for(j=1;j<=num[i];j++)
{
for(k=sum;k>=val[i];k--)
{
dp[k]=max(dp[k],dp[k-val[i]]+val[i]);
}
}
}
printf("%d %d\n",cnt-dp[sum],dp[sum]);
}
return 0;
}