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(模板题)poj 3074 Sudoku(DLX算法)

2017-04-11 10:03 423 查看
Sudoku

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9871 Accepted: 3580
Description

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

.2738..1.
.1...6735
.......29
3.5692.8.
.........
.6.1745.3
64.......
9518...7.
.8..6534.
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
提示

题意：

给出一组数独状态，你需要在'.'上填上数字来完成数独游戏。

思路：

可以将数独转化为精确覆盖问题，本文的源代码以及转化思想：http://blog.csdn.net/weiguang_123/article/details/7935003，思路也可参考：http://blog.csdn.net/bl0ss0m/article/details/17918705

精确覆盖问题的讲解和DLX算法详解：http://www.cnblogs.com/grenet/p/3145800.html，刘汝佳的算法竞赛入门经典训练指南p406也有提到过。

也可利用正规解数独的方法来DFS求解：http://blog.csdn.net/u011932355/article/details/45849039

示例程序

#include <cstdio>
#include <cstring>
using namespace std;
int sum[325],l[237250],r[237250],u[237250],d[237250],c[237250],mplist[730][325],o[730],num,ans[10][10];
void remove(int x)
{
int i,i1;
l[r[x]]=l[x];
r[l[x]]=r[x];
for(i=d[x];i!=x;i=d[i])
{
for(i1=r[i];i1!=i;i1=r[i1])
{
u[d[i1]]=u[i1];
d[u[i1]]=d[i1];
sum[c[i1]]--;
}
}
}
void resume(int x)
{
int i,i1;
r[l[x]]=x;
l[r[x]]=x;
for(i=d[x];i!=x;i=d[i])
{
for(i1=r[i];i1!=i;i1=r[i1])
{
u[d[i1]]=i1;
d[u[i1]]=i1;
sum[c[i1]]++;
}
}
}
int dfs()
{
int i,i1,x,min;
if(r[0]==0)
{
return 1;
}
min=1000000007;
for(i=r[0];i!=0;i=r[i])
{
if(sum[i]<min)
{
min=sum[i];
x=i;
}
}
remove(x);
for(i=d[x];i!=x;i=d[i])
{
o[num]=(i-1)/324;
num++;
for(i1=r[i];i1!=i;i1=r[i1])
{
remove(c[i1]);
}
if(dfs()==1)
{
return 1;
}
for(i1=l[i];i1!=i;i1=l[i1])
{
resume(c[i1]);
}
num--;
}
resume(x);
return 0;
}
int build()
{
int i,i1,pre,first,pos;
num=0;
for(i=0;324>i;i++)
{
r[i]=i+1;
l[i+1]=i;
}
r[324]=0;
l[0]=324;
for(i=1;324>=i;i++)
{
pre=i;
sum[i]=0;
for(i1=1;729>=i1;i1++)
{
if(mplist[i1][i]==1)
{
sum[i]++;
pos=i1*324+i;
c[pos]=i;
d[pre]=pos;
u[pos]=pre;
pre=pos;
}
}
u[i]=pre;
d[pre]=i;
if(sum[i]==0)
{
return 0;
}
}
for(i=1;729>=i;i++)
{
pre=-1;
first=-1;
for(i1=1;324>=i1;i1++)
{
if(mplist[i][i1]==1)
{
pos=i*324+i1;
if(pre==-1)
{
first=pos;
}
else
{
r[pre]=pos;
l[pos]=pre;
}
pre=pos;
}
}
if(first!=-1)
{
r[pre]=first;
l[first]=pre;
}
}
return 1;
}
void output()
{
int i,i1,x,y,k,r,t;
for(i=0;num>i;i++)
{
r=o[i];
k=r%9;
if(k==0)
{
k=9;
}
t=(r-k)/9+1;
y=t%9;
if(y==0)
{
y=9;
}
x=(t-y)/9+1;
ans[x][y]=k;
}
for(i=1;9>=i;i++)
{
for(i1=1;9>=i1;i1++)
{
printf("%d",ans[i][i1]);
}
}
printf("\n");
}
int main()
{
int i,i1,i2,top,t;
char m[82];
while(scanf("%s",m)!=EOF&&strcmp(m,"end")!=0)
{
top=0;
memset(mplist,0,sizeof(mplist));
for(i=1;9>=i;i++)
{
for(i1=1;9>=i1;i1++)
{
t=(i-1)*9+i1;
if(m[top]=='.')
{
for(i2=1;9>=i2;i2++)
{
mplist[9*(t-1)+i2][t]=1;
mplist[9*(t-1)+i2][81+(i-1)*9+i2]=1;
mplist[9*(t-1)+i2][162+(i1-1)*9+i2]=1;
mplist[9*(t-1)+i2][243+((i-1)/3*3+(i1+2)/3-1)*9+i2]=1;
}
}
else
{
i2=m[top]-'0';
mplist[9*(t-1)+i2][t]=1;
mplist[9*(t-1)+i2][81+(i-1)*9+i2]=1;
mplist[9*(t-1)+i2][162+(i1-1)*9+i2]=1;
mplist[9*(t-1)+i2][243+((i-1)/3*3+(i1+2)/3-1)*9+i2]=1;
}
top++;
}
}
build();
dfs();
output();
}
return 0;
}
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