dfs, bfs, 图和树问题 （一）

1.图表示

Adjacent list

vector<int> adj[N]

weighted graph

vector<pair<int,int>> adj[N];

adjacent matrix

int adj[N][N];

edge list

vector<pair<int, int>> edges;

weighted edge list

vector<tuple<int, int, int>> edges;

2.图遍历

dfs

vector<int> adj[N];
bool visited[N];
void dfs(int s) {
    if (visited[s]) return;
    visited[s] = true;
    // process node s
    for (auto u: adj[s]) {
        dfs(u);
    }
}

acwing842.排列数字

dfs + visited + 回溯(剪枝 + 恢复现场)

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

#include <bits/stdc++.h>
#define N 7
using namespace std;

int n;
int path[N];
bitset<9> bs {};
void dfs(int k) {
    if (k==n) {
        for (int i=0; i<k && path[i] != 0; i++) { 
            cout << path[i] << " ";
        }
        cout << "\n";
        return;
    } else {
        for (int j=1; j<=n; j++) {
            if (bs[j]) continue;
            bs[j] = true;
            path[k] = j;
            dfs(k+1);
            path[k] = 0;
            bs[j] = false;
        } 
    }
}
int main() {
    cin >> n;
    dfs(0);
}

acwing843.n皇后

https://www.acwing.com/problem/content/845/
对角线的处理 + 行坐标->len

反对角线 -> 同条反对角线上col+row相等
对角线 -> 同个对角线上n-row+col相等（相当于回到截距，但是和坐标不同，col往下是增大，所以要+col）
ttl -> 需要用行坐标代表len(greedy)

#include <bits/stdc++.h>
#define N 20
using namespace std;
int n;
int visited_row[N], visited_col[N], visited_dg[N], visited_udg[N];
set<pair<int, int>> paths;

void dfs(int len) {
    if (len == n) {
        for (int i=0; i<n; i++) {
            for (int j=0; j<n; j++) {
                if (paths.find({i, j}) != paths.end()) {
                    cout << "Q";
                } else 
                cout << ".";
            }
            cout << "\n";
        }
        cout << "\n";
        return;
    }
    else 
    for (int row=len; row<n; row ++ ) {
        for (int col=0; col<n; col++) {
            if (visited_row[row] || visited_col[col] || visited_dg[col+row] || visited_udg[n-row+col] ) continue;
            visited_row[row] = visited_col[col] = visited_dg[col+row] = visited_udg[n-row+col] = true;
            paths.insert({row, col});
            dfs(len+1);
            visited_row[row] = visited_col[col] = visited_dg[col+row] = visited_udg[n-row+col] = false;
            paths.erase({row, col});
        }
    }
}

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

#include <bits/stdc++.h>
#define N 20
using namespace std;
int n;
int visited_col[N], visited_dg[N], visited_udg[N];
set<pair<int, int>> paths;

void dfs(int row) {
    if (row == n) {
        for (int i=0; i<n; i++) {
            for (int j=0; j<n; j++) {
                if (paths.find({i, j}) != paths.end()) {
                    cout << "Q";
                } else 
                cout << ".";
            }
            cout << "\n";
        }
        cout << "\n";
        return;
    }
    else 
    for (int col=0; col<n; col++) {
        if (visited_col[col] || visited_dg[col+row] || visited_udg[n-row+col] ) continue;
        visited_col[col] = visited_dg[col+row] = visited_udg[n-row+col] = true;
        paths.insert({row, col});
        dfs(row+1);
        visited_col[col] = visited_dg[col+row] = visited_udg[n-row+col] = false;
        paths.erase({row, col});
    }
}

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

bfs

queue<int> q;
bool visited[N];

void bfs(int s) {
    if (visited[s]) return;
    visited[s] = true;
         
    while(!q.empty()) {
        int s = q.front();q.pop();
        for (auto u: adj[s]) {
            if (visited[u]) continue;
            visited[u] = true;
            q.push(u);
        }
    }
}

acwing844走迷宫

https://www.acwing.com/problem/content/846/
分层形bfs

#include <bits/stdc++.h>
#define N 101
using namespace std;
int mat[N][N];
int n, m, len;

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

int main() {
    cin >> n >> m;
    for (int i=0; i<n; i++) {
        for (int j=0; j<m; j++) {
            int a;
            cin >> a;
            mat[i][j] = a; 
        }
    }
    
    // bfs();
    queue<pair<int,int>> q {{make_pair(0,0)}};
    len = 0;
    
    while (!q.empty()) {
        int lenq = q.size();
        len++; 
        for (int j=0; j<lenq; j++) {
            auto [x, y] = q.front(); q.pop();
            for (int i=0; i<4; i++) {
                int new_x = x + dx[i], new_y = y+dy[i];
                if (new_x >= 0 && new_x < n && new_y >= 0 && new_y < m && !mat[new_x][new_y]) {
                    mat[new_x][new_y] = 1;
                    if (new_x == n-1 && new_y == m-1) {
                        cout << len;
                        return 0;
                    }
                    q.push({new_x, new_y});
                }
            }
        }
        
    }
} 

acwing845八数码

https://www.acwing.com/problem/content/description/847/
直接暴力bfs + map
向量和矩阵坐标的转换 -> x = idx/3; y=idx%3;
矩阵转回向量坐标 -> idx = x*3 + y;

#include <bits/stdc++.h>
using namespace std;
unordered_map<string, int> str_dist {};
string endstr {"12345678x"};
int x_idx, dx[4]{0,1,0,-1},dy[4]{1,0,-1,0};

int bfs(string str) {
    if (endstr == str) return 0;    
    int dist=0;
    queue<pair<string, int>> q{{make_pair(str, x_idx)}};
    
    while(q.size()) {
        int lenQ = q.size();dist++;
        
        for (int i=0; i<lenQ; i++) {
            auto [cur_str, idx] = q.front(); q.pop();
            int x=idx/3, y=idx%3;
            for (int j=0; j<4; j++) {
                int new_x = x+dx[j], new_y=y+dy[j];
                if (new_x>=0 && new_x<3 && new_y<3 && new_y>=0) {
                    swap(cur_str[idx], cur_str[new_x*3+new_y]);
                    if (str_dist.find(cur_str) == str_dist.end()) {
                        if (cur_str == endstr) return dist;
                        str_dist[cur_str] = dist;
                        q.push(make_pair(cur_str, new_x*3+new_y));
                    }
                    swap(cur_str[idx], cur_str[new_x*3+new_y]);
                }
            }
        }
    }
    return -1;
}

int main() {
    string str;
    for (int i=0; i<9; i++) {
        char a;
        cin >> a;
        if (a=='x') x_idx= i; 
        str += a;
    }
    
    cout << bfs(move(str));
}

康托展开优化版

应用

• Connectivity check
• Finding cycles
• Bipartiteness check（两色法）

拓扑排序

Shortest paths

unweighted graph - bfs

Bellman–Ford

Dijkstra

Floyd–Warshall

最大最小流问题

最大流