HDU 4069 squigugly Sudoku

HDU_4069

I remember the day before when I was discussing the Data independence issue in the group, I realized that I had another dancing links thing. As a result, the second day of the Fuzhou online competition had a Data independence issue ......

After learning the Dancing Links, I finally dropped this question A. the difference between this question and the ordinary Sudoku lies in the grouping problem. In the past, it was A small palace that could not repeat 1-9, this time, the shape is irregular. We need to first split the graph into nine groups for further processing.

#include<stdio.h>
#include<string.h>
#define INF 1000000000
const int N = 9;
const int mn = N * N * N * 4  + N * N * 4 + N;
const int nn = N * N * 4 + N;
int U[mn], D[mn], L[mn], R[mn], C[mn], H[mn], X[mn];
int S[nn], Q[nn], visc[nn], size;
int a[nn][nn], ans[nn][nn], color[nn][nn], col, vis[nn][nn], num;
int dx[] = {-1, 1, 0, 0}, dy[] = {0 , 0, -1, 1}, wall[] = {16, 64, 128, 32};
void dfs(int x, int y)
{
int i, newx, newy;
for(i = 0; i < 4; i ++)
    {
        newx = x + dx[i];
        newy = y + dy[i];
if(newx >= 0 && newx < N && newy >= 0 && newy < N
        && !vis[newx][newy] && !(wall[i] & a[x][y]))
        {
            vis[newx][newy] = 1;
            color[newx][newy] = col ++;
            dfs(newx, newy);
        }
    }
}
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 + (color[i][j] / 10) * 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] = L[size] = R[size] = size;
else
    {
        R[size] = R[H[r]];
        L[R[H[r]]] = size;
        L[size] = H[r];
        R[H[r]] = size;
    }
}
void init()
{
int i, j, k, r, c1, c2, c3, c4;
for(i = 0; i < N; i ++)
for(j = 0; j < N; j ++)
            scanf("%d", &a[i][j]);
    memset(vis, 0, sizeof(vis));
    col = 0;
for(i = 0; i < N; i ++)
for(j = 0; j < N; j ++)
if(!vis[i][j])
            {
                vis[i][j] = 1;
                color[i][j] = col ++;
                dfs(i , j);
                col ++;
            }
    prepare(mn, N * N * 4);
    memset(visc, 0, sizeof(visc));
for(i = 0; i < N; i ++)
for(j = 0; j < N; j ++)
if(a[i][j] & 15)
            {
                place(r, c1, c2, c3, c4, i, j, a[i][j] & 15);
                link(r, c1), link(r, c2), link(r, c3), link(r, c4);
                visc[c2] = visc[c3] = visc[c4] = 1;
            }
for(i = 0; i < N; i ++)
for(j = 0; j < N; j ++)
if(!(a[i][j] & 15))
for(k = 1; k <= N; k ++)
                {
                    place(r, c1, c2, c3, c4, i, j, k);
if(visc[c2] || visc[c3] || visc[c4])
continue;
                    link(r, c1), link(r, c2), link(r, c3), link(r, c4);
                }
}
void remove(int c)
{
int i, j;
    R[L[c]] = R[c];
    L[R[c]] = L[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])
        {
            S[C[j]] ++;
            U[D[j]] = j;
            D[U[j]] = j;
        }
    L[R[c]] = c;
    R[L[c]] = c;
}
int dance(int cur)
{
int i, j, c,temp;
if(!R[0])
    {
        num ++;
if(num == 2)
return 1;
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 0;
    }
    temp = INF;
for(i = R[0]; i != 0; i = R[i])
if(S[i] < temp)
        {
            c = i;
            temp = S[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;
}
void printresult()
{
int i, j;
for(i = 0; i < N; i ++)
    {
for(j = 0; j < N; j ++)
            printf("%d", ans[i][j]);
        printf("\n");
    }
}
int main()
{
int t, tt;
    scanf("%d", &t);
for(tt = 0; tt < t; tt ++)
    {
        init();
        printf("Case %d:\n", tt + 1);
        num = 0;
if(dance(0))
            printf("Multiple Solutions\n");
else if(num == 1)
            printresult();
else
            printf("No solution\n");
    }
return 0;
}