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  • hdu1426 Sudoku Killer

    简单的DFS,将所有未填的位置依次保存起来,之后,对每一个未填的位置依次枚举,往下搜即可

    ;;

    本来想用DLX的做的,hdu3111是过了,可是原先那个模板在这道题里就是超时,不解啊,换一个模板就过了、

    DFS版

    #include<stdio.h>
    struct point
    {
    int x, y;
    } p[81];
    int num, flag, map[10][10];
    int judge(int n, int k)
    {
    int i, j, x, y;
    for(i = 0; i < 9; i++)
    {
    if(i != p[n].y && map[p[n].x][i] == k)
    return 0;
    if(i != p[n].x && map[i][p[n].y] == k)
    return 0;
    }
    x = p[n].x / 3 * 3;
    y = p[n].y / 3 * 3;
    for(i = 0; i < 3; i++)
    for(j = 0; j < 3; j++)
    if(map[x+i][y+j] == k && (i != p[n].x || j != p[n].y))
    return 0;
    return 1;
    }
    void dfs(int n)
    {
    int i;
    if(n == num)
    {
    flag = 1;
    return;
    }
    for(i = 1; i < 10; i++)
    {
    if(judge(n, i))
    {
    map[p[n].x][p[n].y] = i;
    dfs(n + 1);
    if(flag)
    return;
    map[p[n].x][p[n].y] = 0;
    }
    }
    return;
    }
    int main()
    {
    int i, j, t = 0;
    char s[2];
    while(scanf("%s", s) != -1)
    {
    num = flag = 0;
    if(s[0] == '?')
    p[num].x = 0, p[num++].y = 0, map[0][0] = 0;
    else
    map[0][0] = s[0] - '0';
    for(i = 0; i < 9; i++)
    for(j = 0; j < 9; j++)
    if(i || j)
    {
    scanf("%s", s);
    if(s[0] == '?')
    p[num].x = i, p[num++].y = j, map[i][j] = 0;
    else
    map[i][j] = s[0] - '0';
    }
    dfs(0);
    if(t++)
    printf("\n");
    for(i = 0; i < 9; i++)
    {
    for(j = 0; j < 8; j++)
    printf("%d ", map[i][j]);
    printf("%d\n", map[i][8]);
    }
    }
    return 0;
    }
    DLX
    #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;
    }



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  • 原文地址:https://www.cnblogs.com/nanke/p/2363678.html
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