(中等) HDU 4979 A simple math problem. , DLX+重复覆盖+打表。

　　Description

　　Dragon loves lottery, he will try his luck every week. One day, the lottery company brings out a new form of lottery called accumulated lottery. In a normal lottery, you pick 7 numbers from N numbers. You will get reward according to how many numbers you match. If you match all 7 numbers, you will get the top prize for 1 billion dollars!!! Unlike normal lottery, an M-accumulated lottery allows you to pick M numbers from N numbers. If M is big enough, this may significantly increase your possibility to win. (Of course it cost more…)

　　Some people buy multiple accumulated lotteries to guarantee a higher possibility to get the top prize. Despite of this, it’s still not worthy to guarantee a top prize.Knowing this, Dragon changes his target to second tier prize. To get a second tier prize, you need to contain all of the R numbers with M numbers picked.Given N, M and R, Dragon wants to know how many M-accumulated lotteries he needs to buy, so that he can guarantee that he can get at least the second tier prize.

　　这个题目就是让我们找到至少需要几个彩票，能够覆盖每一个R。

　　然后构造的话就是C(N,R)列，C(N,M)行。。。。。。

　　不过对于8,5,4这个数据要15分钟才能出答案。。。。。。所以。。。。。。要打表。。。。。。（坑。。。。。。）

　　现在还不明白为啥要这么慢。。。。。。

打表的代码如下：

#include<iostream>
#include<cstring>

using namespace std;

const int MaxN=100;
const int MaxM=100;
const int MaxNode=MaxN*MaxM;
const int INF=10e8;

struct DLX
{
    int U[MaxNode],D[MaxNode],L[MaxNode],R[MaxNode],col[MaxNode];
    int H[MaxN],S[MaxM];
    int ans;
    int n,m,size;

    void init(int _n,int _m)
    {
        n=_n;
        m=_m;
        size=m;
        ans=INF;

        for(int i=0;i<=m;++i)
        {
            L[i]=i-1;
            R[i]=i+1;
            U[i]=D[i]=i;

            S[i]=0;
        }
        L[0]=m;
        R[m]=0;

        for(int i=0;i<=n;++i)
            H[i]=-1;
    }

    void Link(int r,int c)
    {
        col[++size]=c;
        ++S[c];

        U[size]=U[c];
        D[size]=c;
        D[U[c]]=size;
        U[c]=size;

        if(H[r]==-1)
            H[r]=L[size]=R[size]=size;
        else
        {
            L[size]=L[H[r]];
            R[size]=H[r];
            R[L[H[r]]]=size;
            L[H[r]]=size;
        }
    }

    void remove(int c)
    {
        for(int i=D[c];i!=c;i=D[i])
        {
            L[R[i]]=L[i];
            R[L[i]]=R[i];
        }
    }

    void resume(int c)
    {
        for(int i=U[c];i!=c;i=U[i])
            L[R[i]]=R[L[i]]=i;
    }

    bool vis1[MaxM];

    int getH()
    {
        int ret=0;

        for(int i=R[0];i!=0;i=R[i])
            vis1[i]=1;

        for(int i=R[0];i!=0;i=R[i])
            if(vis1[i])
            {
                ++ret;
                vis1[i]=0;

                for(int j=D[i];j!=i;j=D[j])
                    for(int k=R[j];k!=j;k=R[k])
                        vis1[col[k]]=0;                                //!!!!!!!!!!!!!!
            }

        return ret;
    }

    void Dance(int d)
    {
        if(getH()+d>=ans)
            return;

        if(R[0]==0)
        {
            if(d<ans)
                ans=d;

            return;
        }

        int c=R[0];

        for(int i=R[0];i!=0;i=R[i])
            if(S[i]<S[c])
                c=i;

        for(int i=D[c];i!=c;i=D[i])
        {
            remove(i);

            for(int j=R[i];j!=i;j=R[j])
                remove(j);

            Dance(d+1);

            for(int j=L[i];j!=i;j=L[j])
                resume(j);

            resume(i);
        }
    }
};

DLX dlx;
int N,M,R;
int C[10][10];

void getC()
{
    for(int i=0;i<=8;++i)
        C[i][0]=1;

    for(int i=1;i<=8;++i)
        for(int j=1;j<=i;++j)
            C[i][j]=C[i-1][j-1]+C[i-1][j];
}

void numTOp(int *s,int x,int n,int m,int d)
{
    if(m<=0)
        return;

    if(x>=C[n-1][m-1])
        numTOp(s,x-C[n-1][m-1],n-1,m,d+1);
    else
    {
        s[0]=d;
        numTOp(s+1,x,n-1,m-1,d+1);
    }
}

int pTOnum(int *s1,int *s2)
{
    int vis[10],ans1=0;

    memset(vis,0,sizeof(vis));

    for(int i=0;i<R;++i)
        vis[s1[s2[i]-1]]=1;

    int tR=R-1;

    for(int i=1;i<=N && tR>=0;++i)
    {
        if(vis[i]==0)
            ans1+=C[N-i][tR];
        else
            --tR;
    }

    return ans1;
}

void slove()
{
    int s[10];
    int ts[10],temp;

    dlx.init(C[N][M],C[N][R]);

    for(int i=0;i<C[N][M];++i)
    {
        numTOp(s,i,N,M,1);

        for(int j=0;j<C[M][R];++j)
        {
            numTOp(ts,j,M,R,1);

            temp=pTOnum(s,ts);

            dlx.Link(i+1,temp+1);
        }
    }

    dlx.Dance(0);

    cout<<dlx.ans<<endl;
}

int main()
{
    ios::sync_with_stdio(false);

    int T;

    cin>>T;

    getC();

    for(int cas=1;cas<=T;++cas)
    {
        cin>>N>>M>>R;

        cout<<"Case #"<<cas<<": ";
        slove();
    }

    return 0;
}

把结果保存到文件里然后再复制到一个数组就行了。。。。。。