#include<iostream>
using namespace std;
const int N = 10;
int path[N]; //保存序列
bool st[N];  //数字是否被用过，标记数组 
int n; 

void dfs(int u)
{
    if(u > n) //数字填完了，输出
    {
        for(int i = 1; i <= n; i++) //输出方案
            cout << path[i] << " ";
        cout << endl;
    }
    else
    {
        for(int i = 1; i <= n; i++) //空位上可以选择的数字为:1 ~ n
        {
            if(!st[i]) //如果数字 i 没有被用过
            {
                path[u] = i; //放入空位
                st[i] = true; //数字被用，修改状态
                dfs(u + 1); //填下一个位
                //回溯：原路返回，把原来满足条件的位置的状态修改回来
                st[i] = false;
            }
        }
    }
}

int main()
{
    cin >> n;
    dfs(1);
    return 0;
}

#include <iostream>
using namespace std;
const int N = 20;

int n;
char g[N][N];
bool col[N], dg[N * 2], udg[N * 2]; // 对角线数组大小要开2倍

// 按行搜索
void dfs(int y) {
    if (y == n) { // 所有行都已成功放置
        for (int i = 0; i < n; i ++ ) puts(g[i]);
        puts("");
        return;
    }

    // 遍历当前行的每一列
    for (int x = 0; x < n; x ++ ) {
        // 剪枝：判断列、主对角线、副对角线是否冲突。+n是为了保证下标非负
        if (!col[x] && !dg[y - x + n] && !udg[y + x]) {
            g[y][x] = 'Q';
            col[x] = dg[y - x + n] = udg[y + x] = true;
            dfs(y + 1);
            //回溯：产生了一种方案后原路返回，把原来满足条件的位置的状态修改回来
            col[x] = dg[y - x + n] = udg[y + x] = false;
            g[y][x] = '.';
        }
    }
}

int main() {
    cin >> n;
    for (int i = 0; i < n; i ++ ) {
        for (int j = 0; j < n; j ++ ) {
            g[i][j] = '.';
        }
    }
    dfs(0);
    return 0;
}

#include<iostream>
using namespace std;
const int N = 10;
int n;
bool board[N][N];
void init() {
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            board[i][j] = false;
        }
    }
}
bool isValid(int row, int col) {
    for (int i = 0; i < n; i++) {
        if (board[i][col]) {
            return false;
        }
    }
    for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--) {
        if (board[i][j]) {
            return false;
        }
    }
    for (int i = row - 1, j = col + 1; i >= 0 && j < n; i--, j++) {
        if (board[i][j]) {
            return false;
        }
    }
    return true;
}
void backtracking(int row) {
    if (row == n) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j]) {
                    printf("Q");
                } else {
                    printf(".");
                }
            }
            printf("\n");
        }
        printf("\n");
        return;
    }
    for (int i = 0; i < n; i++) {
        if (isValid(row, i)) {
            board[row][i] = true;
            backtracking(row + 1);
            board[row][i] = false;
        }
    }
}
void solve() {
    init();
    backtracking(0);
}
int main() {
    scanf("%d", &n);
    solve();
    return 0;
}

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;

typedef pair<int, int> PII;
const int N = 110;

int n, m;
int g[N][N]; // 迷宫
int d[N][N]; // 距离数组

int bfs() {
    queue<PII> q;
    memset(d, -1, sizeof d);

    q.push({0, 0});
    d = 0;

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

    while (!q.empty()) {
        PII t = q.front();
        q.pop();

        for (int i = 0; i < 4; i++) {
            int x = t.first + dx[i], y = t.second + dy[i];
            // 判断是否越界、可走、未访问
            if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 0 && d[x][y] == -1) {
                d[x][y] = d[t.first][t.second] + 1;
                q.push({x, y});
            }
        }
    }
    return d[n - 1][m - 1];
}

int main() {
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> g[i][j];
        }
    }
    cout << bfs() << endl;
    return 0;
}

#include<iostream>
#include<queue>
using namespace std;
const int N = 1005;
char grid[N][N];
int n, m, srtX, srtY, dstX, dstY;
const int dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

int bfs() {
    int cnt = 0;
    queue<pair<int, int>> que;
    if (grid[srtX][srtY] == '*') {
        return -1;
    }
    que.emplace(srtX, srtY);
    grid[srtX][srtY] = '*'; // 直接修改地图作为访问标记

    while (!que.empty()) {
        int t = que.size(); // 当前层的节点数
        while (t--) {
            auto [x, y] = que.front();
            que.pop();
            if (x == dstX && y == dstY) {
                return cnt;
            }
            for (int i = 0; i < 4; i++) {
                int nx = x + dir[i];
                int ny = y + dir[i];
                if (nx >= 1 && nx <= n && ny >= 1 && ny <= m && grid[nx][ny] == '.') {
                    que.emplace(nx, ny);
                    grid[nx][ny] = '*';
                }
            }
        }
        cnt++; // 层数加一
    }
    return -1;
}

int main() {
    scanf("%d %d", &n, &m);
    scanf("%d %d %d %d", &srtX, &srtY, &dstX, &dstY);
    for (int i = 1; i <= n; i++) {
        scanf("%s", grid[i] + 1);
    }
    printf("%d", bfs());
    return 0;
}

#include <iostream>
#include <string>
#include <queue>
#include <unordered_map>
#include <algorithm>
using namespace std;

int bfs(string start) {
    string end = "12345678x";
    queue<string> q;
    unordered_map<string, int> d;

    q.push(start);
    d[start] = 0;

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

    while (!q.empty()) {
        string t = q.front();
        q.pop();

        if (t == end) return d[t];

        int dist = d[t];
        int k = t.find('x');
        int x = k / 3, y = k % 3; // 字符串下标转二维坐标

        for (int i = 0; i < 4; i++) {
            int a = x + dx[i], b = y + dy[i];
            if (a >= 0 && a < 3 && b >= 0 && b < 3) {
                string next_state = t;
                swap(next_state[k], next_state[a * 3 + b]); // 交换生成新状态
                if (!d.count(next_state)) { // 如果新状态未访问过
                    d[next_state] = dist + 1;
                    q.push(next_state);
                }
            }
        }
    }
    return -1;
}

int main() {
    string start;
    char c;
    for (int i = 0; i < 9; i++) {
        cin >> c;
        start += c;
    }
    cout << bfs(start) << endl;
    return 0;
}

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1e5 + 10, M = N * 2;
int h[N], e[M], ne[M], idx;
bool st[N];
int n;
int ans = N; // 存储重心的最大连通块大小

void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

// 返回以u为根的子树的节点数
int dfs(int u) {
    st[u] = true;
    int sum = 1; // 包含节点u本身
    int res = 0; // 存储删除u后，其子树构成的连通块的最大值

    for (int i = h[u]; i != -1; i = ne[i]) {
        int j = e[i];
        if (!st[j]) {
            int s = dfs(j); // s是子树j的大小
            res = max(res, s);
            sum += s;
        }
    }

    res = max(res, n - sum); // u上方的连通块大小
    ans = min(ans, res); // 更新全局答案

    return sum;
}

int main() {
    cin >> n;
    memset(h, -1, sizeof h);
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        add(a, b), add(b, a);
    }
    dfs(1);
    cout << ans << endl;
    return 0;
}

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;

const int N = 100010;
int h[N], e[N], ne[N], idx;
int d[N]; // 存储距离，-1表示未访问
int n, m;

void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int bfs() {
    memset(d, -1, sizeof d);
    queue<int> q;

    q.push(1);
    d = 0;

    while (!q.empty()) {
        int t = q.front();
        q.pop();

        for (int i = h[t]; i != -1; i = ne[i]) {
            int j = e[i];
            if (d[j] == -1) {
                d[j] = d[t] + 1;
                q.push(j);
            }
        }
    }
    return d[n];
}

int main() {
    cin >> n >> m;
    memset(h, -1, sizeof h);
    while(m--) {
        int a, b;
        cin >> a >> b;
        add(a, b);
    }
    cout << bfs() << endl;
    return 0;
}

#include<iostream>
#include<cstring>
using namespace std;
const int N = 1e5 + 5;
int h[N], e[N], ne[N], d[N];
int q[N];
int idx;
int n, m;

void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int bfs() {
    memset(d, -1, sizeof(d));
    int hh = 0, tt = 0;
    q = 1;
    d = 0;

    while (hh <= tt) {
        int cur = q[hh++];
        for (int i = h[cur]; i != -1; i = ne[i]) {
            int j = e[i];
            if (d[j] == -1) {
                d[j] = d[cur] + 1;
                q[++tt] = j;
            }
        }
    }
    return d[n];
}
int main() {
    idx = 0;
    memset(h, -1, sizeof(h));
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        add(a, b);
    }
    printf("%d", bfs());
    return 0;
}

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 100010;
int h[N], e[N], ne[N], idx;
int q[N]; // 数组模拟队列
int d[N]; // 存储点的入度
int n, m;

void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

bool topsort() {
    int hh = 0, tt = -1;
    // 将所有入度为0的点入队
    for (int i = 1; i <= n; i++) {
        if (d[i] == 0) q[++tt] = i;
    }

    while (hh <= tt) {
        int t = q[hh++];
        for (int i = h[t]; i != -1; i = ne[i]) {
            int j = e[i];
            d[j]--;
            // 如果此点入度-1后为0则入队
            if (d[j] == 0) q[++tt] = j;
        }
    }
    // 如果所有点都入队，说明存在拓扑序列
    return tt == n - 1;
}

int main() {
    cin >> n >> m;
    memset(h, -1, sizeof h);
    while (m--) {
        int a, b;
        cin >> a >> b;
        add(a, b);
        d[b]++;
    }

    if (topsort()) {
        for (int i = 0; i < n; i++) cout << q[i] << " ";
        cout << endl;
    } else {
        cout << -1 << endl;
    }
    return 0;
}

#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int N = 2e5 + 5;
int n, m;
vector<int> adj[N];
int inDegree[N];

void topSort() {
    priority_queue<int, vector<int>, greater<int>> que; // 优先队列保证字典序
    for (int i = 1; i <= n; i++) {
        if (inDegree[i] == 0) {
            que.push(i);
        }
    }

    vector<int> result;
    while (!que.empty()) {
        int cur = que.top();
        que.pop();
        result.push_back(cur);

        for (int neighbor : adj[cur]) {
            inDegree[neighbor]--;
            if (inDegree[neighbor] == 0) {
                que.push(neighbor);
            }
        }
    }

    if (result.size() == n) {
        for (int i = 0; i < n; i++) {
            printf("%d%c", result[i], i == n - 1 ? '\n' : ' ');
        }
    } else {
        printf("Error: The graph has a cycle.\n");
    }
}

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        adj[a].push_back(b);
        inDegree[b]++;
    }
    topSort();
    return 0;
}

#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 510, M = 10010;
const int INF = 0x3f3f3f3f;

struct Edge {
    int a, b, w;
} edges[M];

int d[N];
int last[N]; // 备份数组
int n, m, k;

void bellman_ford() {
    memset(d, 0x3f, sizeof d);
    d = 0;

    // 迭代 k 次
    for (int i = 0; i < k; i++) {
        memcpy(last, d, sizeof d); // 备份上一轮的距离
        // 遍历所有边进行松弛
        for (int j = 0; j < m; j++) {
            auto e = edges[j];
            if (last[e.a] != INF) { // 只有当起点可达时才松弛
                d[e.b] = min(d[e.b], last[e.a] + e.w);
            }
        }
    }
}

int main() {
    cin >> n >> m >> k;
    for (int i = 0; i < m; i++) {
        int a, b, w;
        cin >> a >> b >> w;
        edges[i] = {a, b, w};
    }

    bellman_ford();

    if (d[n] > INF / 2) puts("impossible");
    else cout << d[n] << endl;

    return 0;
}