DFS
AcWing 842. 排列数字
#include <iostream>

using namespace std;

const int N = 10;

int n;
int path[N];
bool st[N];

void dfs(int u)
{
    if (u == n)
    {
        for (int i = 0; i < n; i++) cout << path[i] << ' ';
        puts("");
        return;
    }

    for (int i = 1; i <= n; i++)
    {
        if (!st[i])
        {
            path[u] = i;
            st[i] = true;
            dfs(u + 1);
            //恢复现场
            //path[u] = 0;
            st[i] = false;
        }
    }
}

int main()
{
    cin >> n;

    dfs(0);

    return 0;
}
AcWing 843. n-皇后问题
第一种搜索顺序

#include <iostream>

using namespace std;

const int N = 20;

int n;
char map[N][N];
bool col[N], dg[N * 2], udg[N * 2];

void dfs(int u)
{
    if (u == n)   //搜索到n个皇后
    {
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++) cout << map[i][j];
            puts("");
        }
        puts("");
        return;
    }

    for (int i = 0; i < n; i++)
        if (!col[i] && !dg[u + i] && !udg[n - u + i])
        {
            map[u][i] = 'Q';
            col[i] = dg[u + i] = udg[n - u + i] = true;
            dfs(u + 1);
            col[i] = dg[u + i] = udg[n - u + i] = false;
            map[u][i] = '.';
        }   
}

int main()
{
    cin >> n;

    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            map[i][j] = '.';

    dfs(0);

    return 0;
}
第二种搜索顺序

#include <iostream>

using namespace std;

const int N = 20;

int n;
char map[N][N];
bool row[N], col[N], dg[N * 2], udg[N * 2];

void dfs(int x, int y, int s)
{
      //扫描完一行，n-1是一列的最后一个位置，y到达n，就代表一行结束
      if (y == n) y = 0, x++;

      //0 ... n-1 行，当x = n，就代表n行都扫描完了
      if (x == n)
      {
          if (s == n)
          {
              for (int i = 0; i < n; i++) puts(map[i]);
              puts("");
          }
          return;
      }

      //不放
      dfs(x, y + 1, s);

      //放
      if (!row[x] && !col[y] && !dg[x + y] && !udg[x - y + n])
      {
          map[x][y] = 'Q';
          row[x] = col[y] = dg[x + y] = udg[x - y + n] = true;
          dfs(x, y + 1, s + 1);
          row[x] = col[y] = dg[x + y] = udg[x - y + n] = false;
          map[x][y] = '.';
      }
}

int main()
{
    cin >> n;

    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            map[i][j] = '.';

    dfs(0, 0, 0);

    return 0;
}
AcWing 1112. 迷宫
#include <iostream>
#include <cstring>

using namespace std;

const int N = 110;

char g[N][N];
bool st[N][N];

int xa, ya, xb, yb;
int n;

int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

bool dfs(int x, int y)
{   
    if (g[x][y] == '#') return false;   //先要判断这个点是不是"#"
    if (x == ya && y == yb) return true;

    st[x][y] = true;
    for (int i = 0; i < 4; i++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= n) continue;
        if (st[a][b]) continue;

        st[a][b] = true;
        if (dfs(a, b)) return true;
    }

    return false;
}

int main()
{
    int M;
    cin >> M;

    while (M--)
    {
        cin >> n;

        for (int i = 0; i < n; i++) cin >> g[i];
        memset(st, false, sizeof st);

        cin  >> xa >> xb >> ya >> yb;

        if (dfs(xa, xb)) cout << "YES" << endl;
        else cout << "NO" << endl;
    }

    return 0;
}
AcWing 1113. 红与黑
#include <iostream>
#include <cstring>

using namespace std;

const int N = 25;

char g[N][N];
bool st[N][N];
int n, m;

int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

int dfs(int x, int y)
{
    int cnt = 1;

    st[x][y] = true;
    for (int i = 0; i < 4; i++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (g[a][b] != '.') continue;
        if (st[a][b]) continue;

        cnt += dfs(a, b);
    }

    return cnt;
}

int main()
{
    while (cin >> m >> n, n || m)
    {
        for (int i = 0; i < n; i++) cin >> g[i];

        memset(st, 0, sizeof st);

        int x, y;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (g[i][j] == '@')
                    x = i, y = j;

        cout << dfs(x, y) << endl;
    }

    return 0;
}
AcWing 1116. 马走日
#include <iostream>
#include <cstring>

using namespace std;

const int N = 15;

int g[N][N];
bool st[N][N];
int ans;
int n, m, x, y;

int dx[] = {-2, -1, 1, 2, 2, 1, -1, -2};
int dy[] = {1, 2, 2, 1, -1, -2, -2, -1};

void dfs(int x, int y, int cnt)
{
    if (cnt == n * m)
    {
        ans++;
        return ;
    }

    st[x][y] = 1;
    for (int i = 0; i < 8; i++)
    {
        int a = x + dx[i], b = y + dy[i];
        if (a < 0 || a >= n || b < 0 || b >= m) continue;
        if (st[a][b]) continue;

        st[a][b] = 1;
        dfs(a, b, cnt + 1);
        st[a][b] = 0;
    }
}

int main()
{
    int M;
    cin >> M;

    while (M--)
    {
        memset(st, 0, sizeof st);
        cin >> n >> m >> x >> y;
        ans = 0;

        dfs(x, y, 1);
        cout << ans << endl;
    }

    return 0;
}
AcWing 1117. 单词接龙
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 21;

string word[N];
int g[N][N];
int used[N];
int n;
int ans;

void dfs(string dragon, int last)
{
    ans = max((int)dragon.size(), ans);

    used[last]++;

    for (int i = 0; i < n; i++)
        if (g[last][i] && used[i] < 2)
            dfs(dragon + word[i].substr(g[last][i]), i);

    used[last]--;
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++) cin >> word[i];
    char start;
    cin >> start;

    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            string a = word[i], b = word[j];
            for (int k = 1; k < min(a.size(), b.size()); k++)
                if (a.substr(a.size()-k, k) == b.substr(0, k))
                {
                    g[i][j] = k;
                    break;            //两个单词接起来的字符串越短越好
                }
        }

    for (int i = 0; i < n; i++)
        if (word[i][0] == start)
            dfs(word[i], i);  //从word[i]开始搜，当前最后一个单词是i。记一下当前单词是i，判断下一个单词是不是能接

    cout << ans << endl;

    return 0;
}