预处理掉无用状态,二分查找可以切出多少木板。
dfs搜索可行性,加上剪枝优化。
#include <cstdio>
#include <algorithm>
using namespace std;
int n,m,tot,ans,l,r,mid;
int a[55],b[1005],sl[1005],bl[1005];
bool f;
long long sum;
void dfs(int ak,int bk,int w)
{
if(bk==0) f=1;
while(ak<=n&&a[ak]<b[1]) {w+=a[ak];ak++;}
if(f||ak>n)return;
if(w+sl[mid]>sum)return;
int t=ak,t1=ak,t2=bk,t3=w;
if(b[bk]==b[bk+1]&&bk!=mid) t=bl[bk+1];
for(int i=t;i<=n;i++)
if(a[i]>=b[bk])
{
bl[bk]=i;a[i]-=b[bk];
bk--;
dfs(ak,bk,w);
ak=t1;bk=t2;w=t3;a[i]+=b[t2];
}
}
int main()
{
register int i,j;
scanf("%d",&n);
for (i=1;i<=n;i++) scanf("%d",&a[i]);
scanf("%d",&m);
for (i=1;i<=m;i++) scanf("%d",&b[i]);
sort(a+1,a+1+n);
sort(b+1,b+1+m);
while (b[m]>a[n]) m--;
for (i=1;i<=n;i++)
{
if (a[i]>b[1]) a[++tot]=a[i];
}
n=tot;
for (i=1;i<=n;i++) sum+=a[i];
for (i=1;i<=m;i++) sl[i]=sl[i-1]+b[i];
l=1,r=m;
while (l<=r)
{
mid=(l+r)>>1;
f=0;
dfs(1,mid,0);
if(f) ans=mid,l=mid+1;
else r=mid-1;
}
printf("%d\n",ans);
return 0;
}