[SCOI2005] fence (search, two points)

[SCOI2005] fences (luogu)

Solution

There is a greedy strategy - try to meet the needs of small wood

So may first need wood from small to large, and second, how many pieces of wood before the separation of up to meet

There are strategies to reduce the partition between the two - the total length of wood provided must be> = total length of all to meet the board of

How to check it?

Consider thoughts of violence - search, search by planks order to meet the needs of

But certainly need pruning

Pruning 1: Provide the total length of wood - currently discarded length> = all to meet the total length of the plank

Pruning 2: timely abandon not meet any of wooden planks

Code

#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int M=51,N=1001;
int a[M],b[N],n,m,tot,ma,t,sum,s[N],fa[M];
bool dfs(int k,int x,int mid)
{
    if(k<=0) return 1;
    if(sum-t<s[mid]) return 0;
    for(int i=x;i<=m;i++)
        if(fa[i]>=b[k])
        {
            fa[i]-=b[k];
            if(fa[i]<b[1]) t+=fa[i];
            if(b[k]==b[k-1])
            {
                if(dfs(k-1,i,mid)) return 1;
            }
            else if(dfs(k-1,1,mid)) return 1;
            if(fa[i]<b[1]) t-=fa[i];
            fa[i]+=b[k];
        }
    return 0;
}
int main()
{
    scanf("%d",&m);
    for(int i=1;i<=m;i++)
        scanf("%d",&a[i]),ma=max(ma,a[i]);
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&b[++tot]);
        if(b[tot]>ma) tot--;
    }
    n=tot,tot =0;
    sort(b+1,b+1+n);
    for(int i=1;i<=m;i++)
        if(a[i]>=b[1]) a[++tot]=a[i];
    m=tot;
    sort(a+1,a+1+m);
    
    for(int i=1;i<=m;i++) sum+=a[i];
    for(int i=1;i<=n;i++) s[i]=s[i-1]+b[i];
    while(sum<s[n]) n--;
    int l=0,r=n;
    while(l<r)
    {
        t=0;
        for(int i=1;i<=m;i++) fa[i]=a[i];
        int mid=(l+r+1)>>1;
        if(dfs(mid,1,mid)) l=mid;
        else r=mid-1;
    }
    printf("%d\n",l);
    return 0;
}