Acwing-----算法基础课之第三讲搜索与图论（一）

842.排列数字

• 链接：https://www.acwing.com/problem/content/844/

#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);
            st[i] = false;
        }
    }
}

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

843. n-皇后问题

• 链接：https://www.acwing.com/problem/content/845/

#include <iostream>
using namespace std;

const int N = 20;
int n;
char g[N][N];
bool col[N], dg[N], udg[N];

void dfs(int u) {
    if (u == n) {
        for (int i = 0; i < n; ++i) cout << g[i] << endl;
        puts("");
        return;
    }
    
    for (int i = 0; i < n; ++i) {
        if (!col[i] && !dg[u + i] && !udg[n - u + i]) {
            g[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;
            g[u][i] = '.';
        }
    }
}

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 = 20;
int n;
char g[N][N];
bool row[N], col[N], dg[N * 2], udg[N * 2];

void dfs(int x, int y, int s) {
    if (s > n) return;
    if (y == n) y = 0, x++;
    if (x == n) {
        if (s == n) {
            for (int i = 0; i < n; ++i) cout << g[i] << endl;
            puts("");
            
        } 
        return;
    }
    
    g[x][y] = '.';
    dfs(x, y + 1, s);
    
    if (!row[x] && !col[y] && !dg[x + y] && !udg[n - y + x]) {
        row[x] = col[y] = dg[x + y] = udg[ n- y + x] = true;
        g[x][y] = 'Q';
        dfs(x, y + 1, s + 1);
        g[x][y] = '.';
        row[x] = col[y] = dg[x + y] = udg[ n- y + x] = false;
    }
}

int main() {
    cin >> n;

    dfs(0, 0, 0);

    return 0;
}

844. 走迷宫

• 链接：https://www.acwing.com/problem/content/846/

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef pair<int, int> PII;

const int N = 110;

int n, m;
int g[N][N], d[N][N];
PII q[N * N];

int bfs() {
    int hh = 0, tt = 0;
    q[0] = {0, 0};
    
    memset(d, -1, sizeof d);
    d[0][0] = 0;
    
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
    
    while (hh <= tt) {
        auto t = q[hh++];
        for (int i = 0; i < 4; ++i) {
            int x = t.first + dx[i], y = t.second + dy[i];
            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[++tt] = {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;
}

846. 树的重心

• 链接：https://www.acwing.com/problem/content/848/

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 100010, M = N * 2;
int n, m, ans = N;
int h[N], e[M], ne[M], idx;
bool st[N];

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

// 以u为根的子树中点的数量
int dfs(int u) {
    st[u] = true;
    
    // sum:当前子树大小
    // res:每一个连通块最大值
    int sum = 1, res = 0;
    for (int i = h[u]; i != -1; i = ne[i]) {
        int j = e[i];
        if (!st[j]) {
            int s = dfs(j);
            res = max(res, s);
            sum += s;
        }
    }
    res = max(res, n - sum);
    ans = min(ans, res);
    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;
}

847. 图中点的层次

• 链接：https://www.acwing.com/problem/content/849/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
const int N = 100010;
int n, m;
int h[N], e[N], ne[N], d[N], q[N], idx;

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;
    d[1] = 0;
    q.push(1);
    
    while (q.size()) {
        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;
}

848. 有向图的拓扑序列

• 链接：https://www.acwing.com/problem/content/850/

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
const int N = 100010;
int n, m;
int h[N], e[N], ne[N], q[N], d[N], idx;

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

bool topsort() {
    int hh = 0, tt = -1;
    for (int i = 1; i <= n; ++i) {
        if (!d[i]) 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]--;
            if (d[j] == 0) q[++tt] = j;
        }
    }
    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] << " ";
        puts("");
    } else puts("-1");
    return 0;
}