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【BZOJ1082】[SCOI2005]柵欄(搜索)

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【BZOJ1082】[SCOI2005]柵欄(搜索)

題面

BZOJ
洛谷

題解

隨便寫個爆搜,洛谷上就\(80\)分了。先放爆搜代碼:

#include<iostream>
#include<cstdio>
using namespace std;
inline int read()
{
    int x=0;bool t=false;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')t=true,ch=getchar();
    while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
    return t?-x:x;
}
int n,m,a[55],b[1010],l[55],ans;
void dfs(int x,int s)
{
    if(s+n-x+1<=ans)return;
    if(x==n+1){ans=s;return;}
    dfs(x+1,s);
    for(int i=1;i<=m;++i)
        if(l[i]+b[x]<=a[i])
            l[i]+=b[x],dfs(x+1,s+1),l[i]-=b[x];
}
int main()
{
    m=read();
    for(int i=1;i<=m;++i)a[i]=read();
    n=read();
    for(int i=1;i<=n;++i)b[i]=read();
    dfs(1,0);
    printf("%d\n",ans);
}

然後發現可以二分,二分之後再玄學剪枝就好了。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
inline int read()
{
    int x=0;bool t=false;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')t=true,ch=getchar();
    while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
    return t?-x:x;
}
int n,m,a[55],b[1010],sum,tot,s[1010];
bool dfs(int x,int lst)
{
    if(!x)return true;
    if(sum+s[x]>tot)return false;
    for(int i=lst;i<=m;++i)
        if(b[x]<=a[i])
        {
            a[i]-=b[x];if(a[i]<b[1])sum+=a[i];
            bool fl=dfs(x-1,b[x]==b[x-1]?i:1);
            if(a[i]<b[1])sum-=a[i];a[i]+=b[x];
            if(fl)return true;
        }
    return false;
}
int main()
{
    m=read();for(int i=1;i<=m;++i)tot+=(a[i]=read());
    n=read();for(int i=1;i<=n;++i)b[i]=read();
    sort(&a[1],&a[m+1]);sort(&b[1],&b[n+1]);
    for(int i=1;i<=n;++i)s[i]=s[i-1]+b[i];
    while(s[n]>tot)--n;
    int l=0,r=n,ret=0;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(dfs(mid,1))l=mid+1,ret=mid;
        else r=mid-1;   
    }
    printf("%d\n",ret);return 0;
}

【BZOJ1082】[SCOI2005]柵欄(搜索)