Squiggly Sudoku HDU

Today we play a squiggly sudoku, The objective is to fill a 9*9 grid with digits so that each column, each row, and each of the nine Connecting-sub-grids that compose the grid contains all of the digits from 1 to 9. 
Left figure is the puzzle and right figure is one solution. 

Now, give you the information of the puzzle, please tell me is there no solution or multiple solution or one solution.

InputThe first line is a number T(1<=T<=2500), represents the number of case. The next T blocks follow each indicates a case. 
Each case contains nine lines, Each line contains nine integers. 
Each module number tells the information of the gird and is the sum of up to five integers: 
0~9: '0' means this gird is empty, '1' - '9' means the gird is already filled in. 
16: wall to the up 
32: wall to the right 
64: wall to the down 
128: wall to the left 
I promise there must be nine Connecting-sub-grids, and each contains nine girds.OutputFor each case, if there are Multiple Solutions or no solution just output "Multiple Solutions" or "No solution". Else output the exclusive solution.(as shown in the sample output)Sample Input

3
144 18 112 208 80 25 54 144 48
135 38 147 80 121 128 97 130 32
137 32 160 144 114 167 208 0 32
192 100 160 160 208 96 183 192 101
209 80 39 192 86 48 136 80 114
152 48 226 144 112 160 160 149 48
128 0 112 166 215 96 160 128 41
128 39 153 32 209 80 101 136 35
192 96 200 67 80 112 208 68 96 

144 48 144 81 81 16 53 144 48
128 96 224 144 48 128 103 128 38
163 208 80 0 37 224 209 0 32
135 48 176 192 64 112 176 192 104
192 101 128 89 80 82 32 150 48
149 48 224 208 16 48 224 192 33
128 0 114 176 135 0 80 112 169
137 32 148 32 192 96 176 144 32
192 96 193 64 80 80 96 192 96

144 88 48 217 16 16 80 112 176
224 176 129 48 128 40 208 16 37
145 32 128 96 196 96 176 136 32
192 32 227 176 144 80 96 192 32
176 192 80 98 160 145 80 48 224
128 48 144 80 96 224 183 128 48
128 36 224 144 51 144 32 128 105
131 64 112 136 32 192 36 224 176
224 208 80 64 64 116 192 83 96

Sample Output

Case 1:
521439678
763895124
984527361
346182795
157964832
812743956
235678419
479216583
698351247
Case 2:
No solution
Case 3:
Multiple Solutions

题意：一个数独求解，与一般数独不同的是此数独用墙把81个数分成了9 个宫，每个宫都有1到9 。样例中每个数独的数最多是由四个墙和本身的数组成
对于输出，数独无解则输出无解，有多个解则输出多个解，右一个解则输出补全的数独。

思路：与一般的数独解法大致相同，不同的是此题首先用搜索将数独中的数分好宫，然后是当有解时，保存答案。

代码：

#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <iostream>
#define mod 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=10240;
struct DLX
{
    int n,id;
    int L[maxn],R[maxn],U[maxn],D[maxn];
    int C[maxn],S[maxn],loc[maxn][3];//C代表列，S代表每列有的数字数量，loc代表这个数在数独中的位置和数值
    int H[maxn],ans[maxn],ansed;//ansed表示解的个数，ans记录答案
    int fans[maxn],fansed;//复制的ansed,ans数组
    void init(int nn)
    {
        n=nn;
        for(int i=0;i<=n;i++)
        {
            U[i]=D[i]=i;
            L[i]=i-1;
            R[i]=i+1;
        }
        L[0]=n; R[n]=0;
        id=n;
        ansed=0;
        fansed=0;
        memset(S,0,sizeof(S));
        memset(H,-1,sizeof(H));
    }
    void Link(int x,int y,int px,int py,int k)
    {
        ++id;
        D[id]=y; U[id]=U[y];
        D[U[y]]=id; U[y]=id;
        loc[id][0]=px,loc[id][1]=py,loc[id][2]=k;//存放数的位置和数
        C[id]=y;
        S[y]++;//此列1的数量加一
        if(H[x]==-1) 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;
    }
    void fuzhi()//复制已经得到的数独，
    {
        fansed=ansed;
        for(int i=0;i<81;i++)
        {
            fans[i]=ans[i];
        }
    }
    void  dfs(int step)
    {
        if(R[0]==0&&step==81)
        {
            ansed++;
            fuzhi();
            return ;
        }
        int c=R[0];
        for(int i=R[0];i;i=R[i])//优先循环1的数量少的一列
        {
             if(S[i]<S[c])
             { 
                 c=i;
             }
        }
        Remove(c);
        for(int i=D[c];i!=c;i=D[i])
        {
            ans[step]=i;
            for(int j=R[i];j!=i;j=R[j]) Remove(C[j]);
            dfs(step+1);
            if(ansed>=2)//当解多于两个时已是多解，不需要再循环了
            {
                return ;
            }
            for(int j=L[i];j!=i;j=L[j]) Resume(C[j]);
        }
        Resume(c);
    }
}dlx;
struct node
{
    int fxi[4],gong;//fxi表示四个方向，gong表示所在的第几宫
    int x,y,num;//x,y表示行列，num表示数
};
node sd[10][10];
int bj[10][10];
int fx[4]={0,1,0,-1},fy[4]={-1,0,1,0};
void bfs(int x,int y,int z)//搜索同一宫的位置
{
    queue<node> qu;
    node no=sd[x][y];
    bj[x][y]=1;
    qu.push(no);
    while(!qu.empty())
    {
        no=qu.front();
        qu.pop();
        for(int k=0;k<4;k++)
        {
            int i=no.x+fx[k];
            int j=no.y+fy[k];
            if(!no.fxi[k]&&bj[i][j]==0)//没有墙并且没走过
            {
                sd[i][j].gong=z;
                bj[i][j]=1;
                node s=sd[i][j];
                qu.push(s);
            }
        }
    }
}
int main()
{
    int t,ans=0;//ans表示第几个测试任务
    scanf("%d",&t);
    while(t--)
    {
        ans++;
        memset(sd,0,sizeof(sd));
        int js=0;
        dlx.init(81*4);
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            {
                scanf("%d",&sd[i][j].num);
                sd[i][j].x=i;
                sd[i][j].y=j;
                if(sd[i][j].num>=128)//左
                {
                    sd[i][j].num-=128;
                    sd[i][j].fxi[0]=1;
                }
                 if(sd[i][j].num>=64)//下
                {
                    sd[i][j].num-=64;
                    sd[i][j].fxi[1]=1;
                }
                 if(sd[i][j].num>=32)//右
                {
                    sd[i][j].num-=32;
                    sd[i][j].fxi[2]=1;
                }
              if(sd[i][j].num>=16)//上
                {
                    sd[i][j].num-=16;
                    sd[i][j].fxi[3]=1;
                }
            }
        }
        js=0;
        memset(bj,0,sizeof(bj));
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            {
                if(bj[i][j]==0)//循环所有数独
                {
                    bfs(i,j,js);
                    sd[i][j].gong=js;
                    js++;
                }
            }
        }
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            {
                int a,b,c,d;
                if(sd[i][j].num==0)
                {
                    for(int k=1;k<=9;k++)
                    {
                        a=i*9+j+1;
                        b=i*9+k+81;
                        c=j*9+k+81+81;
                        d=sd[i][j].gong*9+k+81+81+81;
                        js++;
                        dlx.Link(js,a,i,j,k);
                        dlx.Link(js,b,i,j,k);
                        dlx.Link(js,c,i,j,k);
                        dlx.Link(js,d,i,j,k);
                    }
                }
                else
                {
                    int k=sd[i][j].num;
                    a=i*9+j+1;
                    b=i*9+k+81;
                    c=j*9+k+81+81;
                    d=sd[i][j].gong*9+k+81+81+81;
                    js++;
                    dlx.Link(js,a,i,j,k);
                    dlx.Link(js,b,i,j,k);
                    dlx.Link(js,c,i,j,k);
                    dlx.Link(js,d,i,j,k);
                }
            }
        }
        dlx.dfs(0);
        int num=dlx.fansed;
        if(num==0)//无解
        {
            printf("Case %d:
No solution
",ans);
        }
        else if(num>1)//多解
        {
            printf("Case %d:
Multiple Solutions
",ans);
        }
        else//一个解
        {
            int no[10][10];
            for(int i=0;i<81;i++)
            {
                int a=dlx.fans[i];
                int x=dlx.loc[a][0];
                int y=dlx.loc[a][1];
                int k=dlx.loc[a][2];
                no[x][y]=k;
            }
            printf("Case %d:
",ans);
            for(int i=0;i<9;i++)
            {
                for(int j=0;j<9;j++)
                {
                    printf("%d",no[i][j]);
                }
                printf("
");
            }
        }
    }
}