Spoj1771-Yet Another N-Queen Problem(精确覆盖)

Description

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

题意：N皇后问题，但是棋盘已经有一些皇后，然后问如何选择皇后的位置使得他们互不攻击，N<=50

解析:N这么大，显然不可能爆搜，这里用到的是精确覆盖DLX算法，行表示每个点，DLX的列对应棋盘的行，列和两个方向的斜对角。

搜到N步就可以了，考虑行就好了，不用考虑斜对角什么的。

代码：

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int INF=1e9+;
const int ms=*;
const int maxn=ms*;
int N,ans[],res[];//ans存储第几个选择的编号，res保存第几行答案是第几列
struct DLX
{
    int n,id;
    int L[maxn],R[maxn],U[maxn],D[maxn];
    int C[maxn],S[maxn],loc[maxn][];
    int H[ms];
    void init(int nn=) //传列长
    {
        n=nn;
        for(int i=;i<=n;i++) U[i]=D[i]=i,L[i]=i-,R[i]=i+;
        L[]=n; R[n]=;
        id=n;
        memset(S,,sizeof(S));
        memset(H,-,sizeof(H));
    }
    void Link(int x,int y)
    {
        ++id;
        D[id]=y; U[id]=U[y];
        D[U[y]]=id; U[y]=id;
        loc[id][]=x,loc[id][]=y;
        C[id]=y; S[y]++;
        if(H[x]==-) H[x]=L[id]=R[id]=id;
        else
        {
            int a=H[x];
            int b=R[a];
            L[id]=a; R[a]=id;
            R[id]=b; L[b]=id;
            H[x]=id;
        }
    }
    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)
    {
        for(int i=U[c];i!=c;i=U[i])
            for(int j=R[i];j!=i;j=R[j])
        {
            S[C[j]]++;
            U[D[j]]=j;
            D[U[j]]=j;
        }
        L[R[c]]=c;
        R[L[c]]=c;
    }
    bool dfs(int step)
    {
        if(step>=N) return true;
        if(R[]==) return false;
        int Min=INF,c=-;
        for(int i=R[];i;i=R[i])
        {
            if(i>N) break;
            if(Min>S[i]){ Min=S[i]; c=i; }
        }
        if(c==-) return false;
        Remove(c);
        for(int i=D[c];i!=c;i=D[i])
        {
            ans[step]=loc[i][];
            for(int j=R[i];j!=i;j=R[j]) Remove(C[j]);
            if(dfs(step+)) return true;
            for(int j=L[i];j!=i;j=L[j]) Resume(C[j]);
        }
        Resume(c);
        return false;
    }
}dlx;
bool vis[*];
int main()
{
    while(scanf("%d",&N)!=EOF)
    {
        dlx.init(N*-);
        memset(vis,false,sizeof(vis));
        int y;
        for(int x=;x<=N;x++)
        {
            scanf("%d",&y);
            if(y==) continue;
            int a=x,b=N+y,c=*N+N+x-y,d=*N+x+y-; //对应的行列斜对角编号
            vis[a]=vis[b]=vis[c]=vis[d]=true; //标记
            int t=(x-)*N+y-;
            dlx.Link(t,a); //连接
            dlx.Link(t,b);
            dlx.Link(t,c);
            dlx.Link(t,d);
        }
        for(int x=;x<=N;x++)
            for(int y=;y<=N;y++)
        {
            int a=x,b=N+y,c=*N+N+x-y,d=*N+x+y-;
            if(vis[a]||vis[b]||vis[c]||vis[d]) continue; //有被占据不考虑
            int t=(x-)*N+y-;
            dlx.Link(t,a);
            dlx.Link(t,b);
            dlx.Link(t,c);
            dlx.Link(t,d);
        }
        if(!dlx.dfs()) printf("No answer find\n");
        else
        {
            for(int i=;i<N;i++) res[ans[i]/N]=ans[i]%N;
            for(int i=;i<N;i++) printf("%d%c",res[i]+,i==N-?'\n':' ');
        }
    }
    return ;
}