学步园

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/4ab7bd00a6c30c2259019eae.html可以下载得到，另外就是Knuth写的《Dancing Links》论文的翻译版。

    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;
}