2024/2/7 图的基础知识

图的存储

B3643 图的存储 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

思路：mp[n][n]用来存邻接矩阵，二维vector用来存每个点连的点

完整代码：

#include 
#define int long long
const int N = 1e5 + 10;
int n, m;
std::vector> g(N);
signed main() {
    std::cin >> n >> m;
    int mp[n + 10][n + 10];
    memset(mp, 0, sizeof(mp));
    for (int i = 0; i < m; i++) {
        int u, v;
        std::cin >> u >> v;
        g[u].push_back(v);//里面装的是u能到的点，u能到v
        g[v].push_back(u);//v能到u
        mp[u][v] = 1;
        mp[v][u] = 1;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            std::cout << mp[i][j] << " ";
        }
        std::cout << "\n";
    }
    for (int i = 1; i <= n; i++) {
        std::sort(g[i].begin(), g[i].end());
    }
    for (int i = 1; i <= n; i++) {
        std::cout << g[i].size() << " ";
        for (int j = 0; j < g[i].size(); j++) {
            std::cout << g[i][j] << " ";
        }
        std::cout << "\n";
    }
    return 0;
}

图的遍历

P3916 图的遍历 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

思路：反向建图，用dfs确定每一个点最大可以到达哪一个点，然后按顺序输出

完整代码：

#include 
#define int long long
const int N = 1e5 + 10;
std::vector> g(N);
int vis[N];
void dfs(int x, int y) {
    if (vis[x] != 0)
        return;
    else {
        vis[x] = y;
        for (int i = 0; i < g[x].size(); i++) {
            dfs(g[x][i], y);
        }
    }
}
signed main() {
    int n, m;
    std::cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        int u, v;
        std::cin >> u >> v;
        g[v].push_back(u);
    }
    for (int i = n; i >= 1; i--) {
        dfs(i, i);
    }
    for (int i = 1; i <= n; i++) {
        std::cout << vis[i] << " ";
    }
    return 0;
}