Sudoku
| Time Limit: 2000MS | Memory Limit: 65536K | |||
| Total Submissions: 13023 | Accepted: 6455 | 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
与POJ 3076一样的方法。
代码：
/* ***********************************************
Author :_rabbit
Created Time :2014/5/1 8:56:15
File Name :F.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
struct DLX{
const static int maxn=20010;
#define FF(i,A,s) for(int i = A[s];i != s;i = A[i])
int L[maxn],R[maxn],U[maxn],D[maxn];
int size,col[maxn],row[maxn],s[maxn],H[maxn];
bool vis[70];
int ans[maxn],cnt;
void init(int m){
for(int i=0;i<=m;i++){
L[i]=i-1;R[i]=i+1;U[i]=D[i]=i;s[i]=0;
}
memset(H,-1,sizeof(H));
L[0]=m;R[m]=0;size=m+1;
}
void link(int r,int c){
U[size]=c;D[size]=D[c];U[D[c]]=size;D[c]=size;
if(H[r]<0)H[r]=L[size]=R[size]=size;
else {
L[size]=H[r];R[size]=R[H[r]];
L[R[H[r]]]=size;R[H[r]]=size;
}
s[c]++;col[size]=c;row[size]=r;size++;
}
void del(int c){//精确覆盖
L[R[c]]=L[c];R[L[c]]=R[c];
FF(i,D,c)FF(j,R,i)U[D[j]]=U[j],D[U[j]]=D[j],--s[col[j]];
}
void add(int c){ //精确覆盖
R[L[c]]=L[R[c]]=c;
FF(i,U,c)FF(j,L,i)++s[col[U[D[j]]=D[U[j]]=j]];
}
bool dfs(int k){//精确覆盖
if(!R[0]){
cnt=k;return 1;
}
int c=R[0];FF(i,R,0)if(s[c]>s[i])c=i;
del(c);
FF(i,D,c){
FF(j,R,i)del(col[j]);
ans[k]=row[i];if(dfs(k+1))return true;
FF(j,L,i)add(col[j]);
}
add(c);
return 0;
}
void remove(int c){//重复覆盖
FF(i,D,c)L[R[i]]=L[i],R[L[i]]=R[i];
}
void resume(int c){//重复覆盖
FF(i,U,c)L[R[i]]=R[L[i]]=i;
}
int A(){//估价函数
int res=0;
memset(vis,0,sizeof(vis));
FF(i,R,0)if(!vis[i]){
res++;vis[i]=1;
FF(j,D,i)FF(k,R,j)vis[col[k]]=1;
}
return res;
}
void dfs(int now,int &ans){//重复覆盖
if(R[0]==0)ans=min(ans,now);
else if(now+A()<ans){
int temp=INF,c;
FF(i,R,0)if(temp>s[i])temp=s[i],c=i;
FF(i,D,c){
remove(i);FF(j,R,i)remove(j);
dfs(now+1,ans);
FF(j,L,i)resume(j);resume(i);
}
}
}
}dlx;
const int SLOT=0;
const int ROW=1;
const int COL=2;
const int SUB=3;
int encode(int a,int b,int c){
return a*81+b*9+c+1;
}
void decode(int code,int &a,int &b,int &c){
code--;
c=code%9;code/=9;
b=code%9;code/=9;
a=code;
}
char str[20][20];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int T;
cin>>T;
while(T--){
for(int i=0;i<9;i++)scanf("%s",str[i]);
dlx.init(324);
for(int r=0;r<9;r++)
for(int c=0;c<9;c++)
for(int k=1;k<=9;k++)
if(str[r][c]==‘0‘||str[r][c]==k+‘0‘){
int p=encode(r,c,k-1);
dlx.link(p,encode(SLOT,r,c));
dlx.link(p,encode(ROW,r,k-1));
dlx.link(p,encode(COL,c,k-1));
dlx.link(p,encode(SUB,(r/3)*3+c/3,k-1));
}
dlx.dfs(0);
// cout<<"jjjdfjfj"<<endl;
//cout<<"hhahha: "<<dlx.cnt<<endl;
for(int i=0;i<dlx.cnt;i++){
int r,c,v;
decode(dlx.ans[i],r,c,v);
// cout<<"pppp"<<endl;
// cout<<"han "<<r<<" "<<c<<" "<<v<<endl;
str[r][c]=v+‘1‘;
}
for(int i=0;i<9;i++)printf("%s\n",str[i]);
}
return 0;
}
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