poj 2676(DLX)

第一次做数独的题目， 最纠结的是建矩阵无疑。。。 想了N久才把这个矩阵给弄好， 真的是很麻烦。 但是确实DLX 好快，解数独几乎是秒杀。。。

Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11405		Accepted: 5645		Special Judge

Description

Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the Figure. In some of the cells are written decimal digits from 1 to 9. The other cells are empty. The goal is to fill the empty cells with decimal digits from 1 to 9, one digit per cell, in such way that in each row, in each column and in each marked 3x3 subsquare, all the digits from 1 to 9 to appear. Write a program to solve a given Sudoku-task. 

Input

The input data will start with the number of the test cases. For each test case, 9 lines follow, corresponding to the rows of the table. On each line a string of exactly 9 decimal digits is given, corresponding to the cells in this line. If a cell is empty it is represented by 0.

Output

For each test case your program should print the solution in the same format as the input data. The empty cells have to be filled according to the rules. If solutions is not unique, then the program may print any one of them.

Sample Input

1
103000509
002109400
000704000
300502006
060000050
700803004
000401000
009205800
804000107

Sample Output

143628579
572139468
986754231
391542786
468917352
725863914
237481695
619275843
854396127

Source

Southeastern Europe 2005

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
#define N 300000
#define INF 0x3fffffff

char g[10][10];
int ans[1000];
int u[5000],d[5000],r[5000],l[5000],num[5000],H[1000],save[5000],save1[5000];
int flag,head;
const int n=729;
const int m=324;
int id;

void prepare()
{
    for(int i=0;i<=m;i++)
    {
        num[i]=0;
        d[i]=i;
        u[i]=i;
        r[i]=i+1;
        l[i+1]=i;
    }
    r[m]=0;
    memset(H,-1,sizeof(H)); // 记录每一行的第一个点
}

void link(int tn,int tm)
{
    id++;
    save1[id]=tn; // 记录行
    ++num[save[id]=tm]; // 记录列
    d[id]=d[tm];
    u[ d[tm] ]=id;
    u[id]=tm;
    d[tm]=id;
    if(H[tn]<0) H[tn]=l[id]=r[id]=id;
    else
    {
        r[id]=r[H[tn]];
        l[ r[H[tn]] ]=id;
        r[ H[tn] ]=id;
        l[id]=H[tn];
    }
}

void build()
{
    id=m;
    int sum;
    prepare();
    int tn=0;
    for(int i=1;i<=81;i++)
    {
        for(int j=1;j<=9;j++)
        {
            ++tn;
            link(tn,i);
        }
    }
    sum=81;
    /////////////////
    for(int i=1;i<=9;i++) // 每一行
    {
        tn=(i-1)*81;
        for(int k=1;k<=9;k++)
        {
            int tk=tn+k;
            for(int j=1;j<=9;j++)
            {
                link(tk,sum+(i-1)*9+k);
                tk+=9;
            }
        }
    }
    sum+=81;
    ///////////////////////
    for(int i=1;i<=9;i++)
    {
        tn=(i-1)*9;
        for(int k=1;k<=9;k++)
        {
            int tk=tn+k;
            for(int j=1;j<=9;j++)
            {
                link(tk,sum+(i-1)*9+k);
                tk+=81;
            }
        }
    }
    sum+=81;
    /////////////////////////
    int tt=0;
    for(int i1=1;i1<=3;i1++)
    {
        for(int j1=1;j1<=3;j1++)
        {
            tn=(i1-1)*81*3+9*3*(j1-1);
            for(int k=1;k<=9;k++)
            {
                ++tt;
                int tk;
                for(int i=1;i<=3;i++)
                {
                    for(int j=1;j<=3;j++)
                    {
                        tk=tn+(i-1)*81+9*(j-1)+k;
                        link(tk,sum+tt);
                    }
                }
            }
        }
    }
}

void remove(int s)
{
    l[ r[s] ]=l[s];
    r[ l[s] ]=r[s];
    for(int i=d[s];i!=s;i=d[i])
        for(int j=r[i];j!=i;j=r[j])
        {
            u[d[j]]=u[j];
            d[u[j]]=d[j];
            num[save[j]]--;
        }
}

void resume(int s)
{
    r[l[s]]=s;
    l[r[s]]=s;
    for(int i=u[s];i!=s;i=u[i])
        for(int j=l[i];j!=i;j=l[j])
        {
            u[d[j]]=j;
            d[u[j]]=j;
            num[save[j]]++;
        }
}

void dfs(int s)
{
    if(flag) return ;
    if(r[head]==head)
    {
        flag=1;
        for(int i=0;i<s;i++)
        {
            int ti,tj,tk;
            int tans=save1[ans[i]]-1;
            ti= (tans)/81+1;
            tj= (tans%81)/9+1;;
            tk= (tans%81)%9+1;
            //printf("<%d %d> ",ti,tj);
            g[ti][tj]=tk+'0';
        }
        return ;
    }
    int mi=INF,tu;
    for(int i=r[head];i!=head;i=r[i])
        if(mi>num[i])
        {
            mi=num[i];
            tu=i;
        }
    remove(tu);
    for(int i=d[tu];i!=tu;i=d[i])
    {
        for(int j=r[i];j!=i;j=r[j])
            remove(save[j]);
        ans[s]=i;
        dfs(s+1);
        for(int j=l[i];j!=i;j=l[j])
            resume(save[j]);
    }
    resume(tu);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        build();
        int tu=0;
        for(int i=1;i<=9;i++)
        {
            for(int j=1;j<=9;j++)
            {
                cin>>g[i][j];
                if(g[i][j]!='0')
                {
                    int kk=g[i][j]-'0';
                    remove( save[ H[tu+kk] ] );
                    for(int i1=r[ H[tu+kk] ];i1 != H[tu+kk];i1=r[i1])
                    {
                        remove( save[i1] );
                    }
                }
                tu+=9;
            }
        }
        flag=0;
        dfs(0);
        printf("\n");
        for(int i=1;i<=9;i++)
        {
            for(int j=1;j<=9;j++)
                printf("%c",g[i][j]);
            printf("\n");
        }
    }
    return 0;
}