poj 3074(DLX)

和之前那道数独一样。。。 搜索很强大。只用了47ms

Sudoku

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7108		Accepted: 2498

Description

In the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids. For example,

.	2	7	3	8	.	.	1	.
.	1	.	.	.	6	7	3	5
.	.	.	.	.	.	.	2	9
3	.	5	6	9	2	.	8	.
.	.	.	.	.	.	.	.	.
.	6	.	1	7	4	5	.	3
6	4	.	.	.	.	.	.	.
9	5	1	8	.	.	.	7	.
.	8	.	.	6	5	3	4	.

Given some of the numbers in the grid, your goal is to determine the remaining numbers such that the numbers 1 through 9 appear exactly once in (1) each of nine 3 × 3 subgrids, (2) each of the nine rows, and (3) each of the nine columns.

Input

The input test file will contain multiple cases. Each test case consists of a single line containing 81 characters, which represent the 81 squares of the Sudoku grid, given one row at a time. Each character is either a digit (from 1 to 9) or a period (used to indicate an unfilled square). You may assume that each puzzle in the input will have exactly one solution. The end-of-file is denoted by a single line containing the word “end”.

Output

For each test case, print a line representing the completed Sudoku puzzle.

Sample Input

.2738..1..1...6735.......293.5692.8...........6.1745.364.......9518...7..8..6534.
......52..8.4......3...9...5.1...6..2..7........3.....6...1..........7.4.......3.
end

Sample Output

527389416819426735436751829375692184194538267268174593643217958951843672782965341
416837529982465371735129468571298643293746185864351297647913852359682714128574936

Source

Stanford Local 2006

#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()
{
    char T[100];
    while(scanf("%s",T))
    {
        if(T[0]=='e') break;
        build();
        int tu=0;
        int tcnt=0;
        for(int i=1;i<=9;i++)
        {
            for(int j=1;j<=9;j++)
            {
                g[i][j]=T[tcnt++];
                if(g[i][j]!='.')
                {
                    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);
        for(int i=1;i<=9;i++)
        {
            for(int j=1;j<=9;j++)
                printf("%c",g[i][j]);
        }
        printf("\n");
    }
    return 0;
}