题目:
给出一个整数集合,求出非空子集中元素和绝对值最小是多少(元素个数尽量少)
题解:
分成两半
爆搜每一半,用map维护前一半的值
每搜出后一半的一个值就去map里找和他和绝对值最小的更新答案
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<map>
typedef long long ll;
using namespace std;
ll n,a[50],ans,tmp,ok;
map <ll,ll> mp;
map <ll,ll> :: iterator it;
ll Abs(ll x){ return x>0?x:-x;}
void DfsFront(ll step,ll state,ll val)
{
if (step>n/2)
{
it=mp.find(val);
if (it!=mp.end())
mp[val]=min(mp[val],state);
else
mp[val]=state;
if (Abs(val)<ans && state!=0)
ans=Abs(val),tmp=state;
if (Abs(val)==ans && state<tmp && state!=0)
tmp=state;
return ;
}
DfsFront(step+1,state,val);
DfsFront(step+1,state+1,val+a[step]);
}
void DfsBack(ll step,ll state,ll val)
{
if (step>n)
{
it=mp.lower_bound(-val);
ll w=it->first;
if (it!=mp.end() && Abs(val+w)<ans && it->second+state!=0)
ans=Abs(val+w),tmp=it->second+state;
if (it!=mp.end() && Abs(val+w)==ans && mp[w]+state<tmp && mp[w]+state!=0)
tmp=mp[w]+state;
if (it!=mp.begin()) it--;
if (it!=mp.end())
{
w=it->first;
if (Abs(val+w)<ans && state+it->second!=0)
ans=Abs(val+w),tmp=it->second+state;
if (Abs(val+w)==ans && it->second+state<tmp && it->second!=0)
tmp=it->second+state;
}
return ;
}
DfsBack(step+1,state,val);
DfsBack(step+1,state+1,val+a[step]);
}
int main()
{
while (scanf("%lld",&n)!=0,n)
{
mp.clear();
ans=mp[0]=0;
for (int i=1;i<=n;i++)
scanf("%lld",&a[i]),ans+=233+Abs(a[i]);
DfsFront(1,0,0);
DfsBack(n/2+1,0,0);
printf("%lld %lld\n",ans,tmp);
}
return 0;
}