usaco4.1.2

usaco4.1.2

太忙没时间啰嗦，二分+dfs
要注意木板有很多大小相同的！！！

其它的自己看看

代码
/*
USER:zyn19961
TASK:fence8
LANG:C++
*/
#include<cstdio>
#include<cstring>
    bool can(int x);
    bool dfs(int now,int lp,int x);
    void sort(int x[],int s,int t);
    int n,a[51],p[51],r;
    int b[1025],sum[1025];
    int ans,tot,waste;
    int main(){
        int s,t,m;
        freopen("fence8.in","r",stdin);
        freopen("fence8.out","w",stdout);
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            tot+=a[i];
            }
        scanf("%d",&r);
        sum[0]=0;
        for(int i=1;i<=r;i++)
            scanf("%d",&b[i]);
        sort(a,1,n);
        sort(b,1,r);
        for(int i=1;i<=r;i++)
            sum[i]=sum[i-1]+b[i];
        s=1;t=r;ans=0;
        while(s<=t){
            m=(s+t+1)/2;
            if(can(m)){
                ans=m;
                s=m+1; 
                }
            else
                t=m-1; 
            }
        printf("%d\n",ans);
        fclose(stdin);
        fclose(stdout);
        return 0;
        } 
    void sort(int x[],int s,int t){
        for(int i=s;i<=t;i++)
            for(int j=i+1;j<=t;j++)
                if(x[i]>x[j]){
                    x[i]^=x[j];
                    x[j]^=x[i];
                    x[i]^=x[j];
                    }
        }
    bool can(int x){
        waste=0;
        for(int i=1;i<=n;i++)
            p[i]=a[i];
        if(dfs(x,0,x))return true;
        else return false; 
        } 
    bool dfs(int now,int lp,int x){//
        int i,j;
        bool change;
        if(now==0)return true;
        if(waste+sum[x]>tot)return false;
        if(now!=x&&b[now]==b[now+1])j=lp;else j=1;//very important
        for(int i=j;i<=n;i++)
            if(p[i]>=b[now]){
                p[i]-=b[now];
                change=false;
                if(p[i]<b[1]){//attention
                    change=true;
                    waste+=p[i];
                    }
                if(dfs(now-1,i,x))return true;
                if(change)waste-=p[i];
                p[i]+=b[now];
                }
        return false;
        }