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[AcWing 166] 数独

作者:互联网

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DFS + 剪枝 + 位运算优化


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#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 9, M = 1 << N;

int ones[M]; // ones[i]表示i的二进制数中1的个数 
int map2[M]; // map2[i]表示log_2(i)
int row[N], col[N], cell[3][3]; // 二进制数,1代表没填数
string str;

// 将三个数组全部设置为未填数字
void init()
{
    for (int i = 0; i < N; i ++)
        row[i] = col[i] = (1 << N) - 1;
    for (int i = 0; i < 3; i ++)
        for (int j = 0; j < 3; j ++)
            cell[i][j] = (1 << N) - 1;
}

// 在(x,y)位置填数字(t+1)
// is_set为true表示填,为false表示不填(用于恢复现场)
void draw(int x, int y, int t, bool is_set)
{
    if (is_set)
        str[x * N + y] = '1' + t;
    else
        str[x * N + y] = '.';
    int v = 1 << t;
    if (!is_set)
        v = -v;
    row[x] -= v;
    col[y] -= v;
    cell[x / 3][y / 3] -= v;
}

int get(int x, int y)
{
    return row[x] & col[y] & cell[x / 3][y / 3];
}

int lowbit(int x)
{
    return x & -x;
}

bool dfs(int cnt)
{
    if (!cnt)
        return true;
    // 找到分支情况最少的格子
    int minv = N + 1;
    int x, y;
    for (int i = 0; i < N; i ++)
        for (int j = 0; j < N; j ++)
            if (str[i * N + j] == '.') {
                int state = get(i, j);
                if (ones[state] < minv) {
                    minv = ones[state];
                    x = i, y = j;
                }
            }
    int state = get(x, y);
    // 在1的位置填数字
    for (int i = state; i; i -= lowbit(i)) {
        int t = map2[lowbit(i)];
        draw(x, y, t, true);
        if (dfs(cnt - 1))
            return true;
        draw(x, y, t, false);
    }
    return false;
}

void solve()
{
    for (int i = 0; i < N; i ++)
        map2[1 << i] = i;
    for (int i = 0; i < 1 << N; i ++)
        for (int j = 0; j < N; j ++)
            ones[i] += i >> j & 1;
    while (cin >> str, str != "end") {
        init();
        // 记录空格子的个数
        int cnt = 0;
        // (i,j)是二维坐标 k是一维坐标
        for (int i = 0, k = 0; i < N; i ++)
            for (int j = 0; j < N; j ++, k ++) {
                if (str[k] != '.') {
                    // 将字符1-9映射到数字0-8
                    int t = str[k] - '1';
                    draw(i, j, t, true);
                }
                else
                    cnt ++;
            }
        dfs(cnt);
        cout << str << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    solve();

    return 0;
}

标签:cnt,int,long,++,坐标,str,166,数独,AcWing
来源: https://www.cnblogs.com/wKingYu/p/16597331.html