[SCOI2005]栅栏 二分+dfs

这个题真的是太nb了，各种骚

二分答案，肯定要减最小的mid个，从大往小搜每一个木板，从大往小枚举所用的木材

当当前木材比最短的木板还短，就扔到垃圾堆里，并记录waste，当 waste+sum>tot 时，return

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define N 2005
using namespace std;
int n,m,w[N],nd[N],ans,ANS,sum[N],tt;
void dfs(int x,int y,int wst,int mid){
    if(wst+sum[mid]>tt)return;
    if(y<=0){ans=1;return;}
    for(int i=x;i<=n;i++){
        if(w[i]>=nd[y]){
            w[i]-=nd[y];
            if(w[i]<nd[1]) wst+=w[i];
            if(nd[y]==nd[y-1]) dfs(i,y-1,wst,mid);
            else dfs(1,y-1,wst,mid);
            if(w[i]<nd[1]) wst-=w[i];
            w[i]+=nd[y];
            if(ans==1)return;
        }
    }
}
bool da(int a,int b){return a>b;}
int main(){
    //freopen("fence8.in","r",stdin);
    //freopen("fence8.out","w",stdout);
    scanf("%d",&n); for(int i=1;i<=n;i++)scanf("%d",&w[i]),tt+=w[i];
    scanf("%d",&m); for(int i=1;i<=m;i++)scanf("%d",&nd[i]);
    sort(w+1,w+n+1,da);sort(nd+1,nd+m+1);
    for(int i=1;i<=m;i++)sum[i]=sum[i-1]+nd[i];
    int l=0,r=m,mid;
    while(l<=r){
        mid=(l+r)>>1;
        ans=0;dfs(1,mid,0,mid);
        if(ans){ANS=mid;l=mid+1;}
        else r=mid-1;
    }
    printf("%d\n",ANS);
    return 0;
}