USACO2.1.4 Healthy Holsteins (holstein)

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对于每一种饲料G(1<=G<=15)，只有选和不选两种，所以直接枚举搜索即可

题目求问最少需多少种，所以dfs深度达到m时，就可以比较后return了

/*ID:xsy97051PROB:holsteinLANG:C++*/#include <iostream>#include <cstdio>#include <cstring> using namespace std;int n,v[26],m,g[16][26],now[26],res[16],dat[16],minn=1000000000; void dfs(int deep,int count){    if(deep==m+1)    {        for(int i=1;i<=n;i++)       if(now[i]<v[i]) return ;        if(count<minn)        {            minn=count;            for(int i=1;i<=minn;i++) res[i]=dat[i];    }    return ;    }         for(int i=1;i<=n;i++)        now[i]+=g[deep][i];             dat[count+1]=deep;         dfs(deep+1,count+1);         for(int i=1;i<=n;i++)        now[i]-=g[deep][i];             dfs(deep+1,count);     return ;} int main(){    freopen("holstein.in","r",stdin);    freopen("holstein.out","w",stdout);     cin>>n;    for(int i=1;i<=n;i++)        cin>>v[i];    cin>>m;    for(int i=1;i<=m;i++)    for(int j=1;j<=n;j++)        cin>>g[i][j];            memset(now,0,sizeof(now));    dfs(1,0);    cout<<minn;    for(int i=1;i<=minn;i++)        cout<<" "<<res[i];cout<<endl;    return 0;}

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