Spoj 1771(Dancing Links 精确覆盖变形)

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本文介绍了一种解决复杂N皇后问题的算法实现，通过将问题转换为DLX精确覆盖问题，并采用特定的数据结构和算法进行求解。详细阐述了算法的设计思路、关键步骤及其实现细节，包括如何通过覆盖行或列来寻找解决方案，最终通过实例演示了算法的有效性和效率。

SPOJ Problem Set (classical)

1771. Yet Another N-Queen Problem

Problem code: NQUEEN

After solving Solution to the n Queens Puzzle by constructing, LoadingTime wants to solve a harder version of the N-Queen Problem. Some queens have been set on particular locations on the board in this problem. Can you help him??

Input

The input contains multiple test cases. Every line begins with an integer N (N<=50), then N integers followed, representing the column number of the queen in each rows. If the number is 0, it means no queen has been set on this row. You can assume there is at least one solution.

Output

For each test case, print a line consists of N numbers separated by spaces, representing the column number of the queen in each row. If there are more than one answer, print any one of them.

Example

Input:
4 0 0 0 0
8 2 0 0 0 4 0 0 0

Output:
2 4 1 3
2 6 1 7 4 8 3 5

Added by:	Bin Jin
Date:	2007-09-06
Time limit:	5s
Source limit:	10000B
Languages:	All except: C++ 4.0.0-8
Resource:	own problem

题目：http://www.spoj.pl/problems/NQUEEN/

分析：这题也可以转化为DLX精确覆盖问题，将棋盘的行列左斜线右斜线作为覆盖点，将每个格子作为操作，但是不同的是只需选折覆盖行，或者覆盖列，之前没有判断，直接扫描全部覆盖，一直无解，找了一个晚上加一个早上阿。。。速度还行，跑了0.5s

代码：

#include<cstdio>
#define mm 777777
#define mn 3333
int U[mm],D[mm],L[mm],R[mm],C[mm],id[mm];
int H[mn],S[mn],Q[mn],P[mn],X[mn],Y[mn];
bool p[mn];
int n,size;
void prepare(int r,int c)
{
    for(int i=0;i<=c;++i)
    {
        S[i]=0;
        U[i]=D[i]=i;
        L[i+1]=i;
        R[i]=i+1;
    }
    R[size=c]=0;
    while(r)H[r--]=-1;
}
void remove(int c)
{
    L[R[c]]=L[c],R[L[c]]=R[c];
    for(int i=D[c];i!=c;i=D[i])
        for(int j=R[i];j!=i;j=R[j])
            U[D[j]]=U[j],D[U[j]]=D[j],--S[C[j]];
}
void resume(int c)
{
    L[R[c]]=R[L[c]]=c;
    for(int i=U[c];i!=c;i=U[i])
        for(int j=L[i];j!=i;j=L[j])
            ++S[C[U[D[j]]=D[U[j]]=j]];
}
bool Dance(int k)
{
    if(k>=n)
    {
        for(int i=0;i<k;++i)P[X[id[Q[i]]]]=Y[id[Q[i]]];
        for(int i=1;i<n;++i)printf("%d ",P[i]);
        printf("%d\n",P[n]);
        return 1;
    }
    if(!R[0])return 0;
    int i,j,c,tmp=mm;
    for(i=R[0];i;i=R[i])
        if(i<=n)
        {
            if(S[i]<tmp)tmp=S[c=i];
        }else break;
    remove(c);
    for(i=D[c];i!=c;i=D[i])
    {
        Q[k]=i;
        for(j=R[i];j!=i;j=R[j])remove(C[j]);
        if(Dance(k+1))return 1;
        for(j=L[i];j!=i;j=L[j])resume(C[j]);
    }
    resume(c);
    return 0;
}
void Link(int r,int c)
{
    ++S[C[++size]=c];
    id[size]=r;
    D[size]=D[c];
    U[D[c]]=size;
    U[size]=c;
    D[c]=size;
    if(H[r]<0)H[r]=L[size]=R[size]=size;
    else
    {
        R[size]=R[H[r]];
        L[R[H[r]]]=size;
        L[size]=H[r];
        R[H[r]]=size;
    }
}
void place(int &p1,int &p2,int &p3,int &p4,int i,int j)
{
    p1=j,p2=n+i,p3=n+n+i-j+n,p4=n+n+n+n-1+i+j-1;
}
int main()
{
    int i,j,k,p1,p2,p3,p4;
    while(scanf("%d",&n)!=-1)
    {
        for(i=1;i<=n*6-2;++i)p[i]=0;
        prepare(n*n,n*6-2);
        for(k=i=1;i<=n;++i)
        {
            scanf("%d",&j);
            if(j)
            {
                place(p1,p2,p3,p4,i,j);
                p[p1]=p[p2]=p[p3]=p[p4]=1;
                Link(k,p1),Link(k,p2),Link(k,p3),Link(k,p4);
                X[k]=i,Y[k]=j;
                ++k;
            }
        }
        for(i=1;i<=n;++i)
            for(j=1;j<=n;++j)
            {
                place(p1,p2,p3,p4,i,j);
                if(p[p1]||p[p2]||p[p3]||p[p4])continue;
                Link(k,p1),Link(k,p2),Link(k,p3),Link(k,p4);
                X[k]=i,Y[k]=j;
                ++k;
            }
        Dance(0);
    }
    return 0;
}