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 ",ANS); return 0; }