HDU 1426 Sudoku Killer
HDU_1426
这个题目貌似是可以用深搜做的，但是我实际上是为了演练Dancing Links才选做的这道数独里的简单题。Dancing Links的基本思想就是把图转化成0-1矩阵并最终化归成选出若干行使得每列有且恰有一个1的问题。
推荐两个文章，一个是momodi写的《Dancing Links在搜索中的应用》，这个在http://wenku.baidu.com/view/4ab7bd
如果你看完这些之后向我一样虽然对算法思想有了大概了解，却不知道代码怎么写，尤其是不知道图应该怎么建，或者说建图的代码怎么写，那么可以看看这个解题报告上的代码http://blog.csdn.net/fp_hzq/article/details/6799640，我觉得这份代码还是写的比较清晰的。
#include<stdio.h>
#include<string.h>
#define INF 1000000000
const int N = 9;
const int mn = N * N * N *(N * N * N * 4) + N;
const int nn = N * N * N + N;
int U[mn], D[mn], L[mn], R[mn], C[mn], X[mn];
int H[nn], Q[nn], S[nn], vis[nn], size;
int a[nn][nn], ans[nn][nn];
void prepare(int r, int c)
{
int i;
for(i = 0; i <= c; i ++)
{
S[i] = 0;
U[i] = D[i] = i;
R[i] = i + 1;
L[i + 1] = i;
}
R[c] = 0;
size = c;
while(r)
H[r --] = -1;
}
void place(int &r, int &c1, int &c2, int &c3, int &c4, int i, int j, int k)
{
r = (i * N + j) * N + k;
c1 = i * N + j + 1;
c2 = N * N + i * N + k;
c3 = 2 * N * N + j * N + k;
c4 = 3 * N * N + ((i / 3) * 3 + j / 3) * N + k;
}
void link(int r, int c)
{
size ++;
C[size] = c;
S[c] ++;
X[size] = r;
D[size] = D[c];
U[D[c]] = size;
U[size] = c;
D[c] = size;
if(H[r] < 0)
{
H[r] = size;
L[size] = size;
R[size] = size;
}
else
{
R[size] = R[H[r]];
L[R[H[r]]] = size;
L[size] = H[r];
R[H[r]] = size;
}
}
void remove(int c)
{
int i, j;
L[R[c]] = L[c];
R[L[c]] = R[c];
for(i = D[c]; i != c; i = D[i])
for(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)
{
int i, j;
for(i = U[c]; i != c; i = U[i])
for(j = L[i]; j != i; j = L[j])
{
U[D[j]] = j;
D[U[j]] = j;
S[C[j]] ++;
}
R[L[c]] = c;
L[R[c]] = c;
}
int dance(int cur)
{
int i, j, c, temp;
if(!R[0])
{
for(i = 0; i < cur ;i ++)
{
int x = (X[Q[i]] - 1) / N / N;
int y = (X[Q[i]] - 1) / N % N;
ans[x][y] = (X[Q[i]] - 1) % N + 1;
}
return 1;
}
temp = INF;
for(i = R[0]; i != 0; i = R[i])
if(S[i] < temp)
{
temp = S[i];
c = i;
}
remove(c);
for(i = D[c]; i != c; i = D[i])
{
Q[cur] = i;
for(j = R[i] ; j != i; j = R[j])
remove(C[j]);
if(dance(cur + 1))
return 1;
for(j = L[i]; j != i; j = L[j])
resume(C[j]);
}
resume(c);
return 0;
}
int init()
{
int i, j, k, r, c1, c2, c3, c4;
char temp[5];
if(scanf("%s", temp) != 1)
return 0;
a[0][0] = (temp[0] == '?' ? 0 : temp[0] - '0');
for(i = 0; i < N; i ++)
for(j = 0; j < N; j ++)
{
if(!i && !j)
continue;
scanf("%s", temp);
a[i][j] = (temp[0] == '?' ? 0 : temp[0] - '0');
}
prepare(N * N * N, N * N * 4);
memset(vis, 0, sizeof(vis));
for(i = 0; i < N; i ++)
for(j = 0; j < N; j ++)
if(a[i][j])
{
place(r, c1, c2, c3, c4, i, j, a[i][j]);
link(r, c1), link(r, c2), link(r, c3), link(r, c4);
vis[c2] = vis[c3] = vis[c4] = 1;
}
for(i = 0; i < N; i ++)
for(j = 0; j < N; j ++)
if(!a[i][j])
for(k = 1; k <= N; k ++)
{
place(r, c1, c2, c3, c4, i, j, k);
if(vis[c2] || vis[c3] || vis[c4])
continue;
link(r, c1), link(r, c2), link(r, c3), link(r, c4);
}
return 1;
}
void printresult()
{
int i, j;
for(i = 0; i < N; i ++)
{
for(j = 0; j < N; j ++)
{
if(j)
printf(" ");
printf("%d", ans[i][j]);
}
printf("\n");
}
}
int main()
{
int t = 0;
while(init())
{
dance(0);
if(t ++)
printf("\n");
printresult();
}
return 0;
}