[jobdu]树的子结构

判断一棵树B是否是A的子树，对A做DFS，然后不断判断是否和B相同。

其实也可以不对A做DFS，直接遍历A中的每个节点和B做树的比较就行了。

#include <iostream>
#include <vector>
using namespace std;

bool sameTree(int a, int b, vector<vector<int> > &treeA, vector<vector<int> > &treeB, vector<int> &valA, vector<int> &valB) {
    if (valA[a] != valB[b]) return false;
    if (treeB[b].size() == 0) return true;
    if (treeA[a].size() != treeB[b].size()) return false;
    for (int i = 0; i < treeA[a].size(); i++) {
        if (!sameTree(treeA[a][i], treeB[b][i], treeA, treeB, valA, valB))
            return false;
    }
    return true;
}

bool subTree(vector<vector<int> > &treeA, vector<vector<int> > &treeB, vector<int> &valA, vector<int> &valB, int root) {
    if (sameTree(root, 1, treeA, treeB, valA, valB)) {
        return true;
    }
    for (int i = 0; i < treeA[root].size(); i++) {
        if (subTree(treeA, treeB, valA, valB, treeA[root][i])) {
            return true;
        }
    }
    return false;
}

int main() {
    int n, m;
    while (cin >> n >> m)
    {
        vector<vector<int> > treeA(n+1);
        vector<vector<int> > treeB(m+1);
        vector<int> valA(n+1);
        vector<int> valB(m+1);
        for (int i = 1; i <= n; i++) {
            int tmp;
            cin >> tmp;
            valA[i] = tmp;
        }
        for (int i = 1; i <= n; i++) {
            int cnt;
            cin >> cnt;
            while (cnt--) {
                int x;
                cin >> x;
                treeA[i].push_back(x);
            }
        }
        for (int i = 1; i <= m; i++) {
            int tmp;
            cin >> tmp;
            valB[i] = tmp;
        }
        for (int i = 1; i <= m; i++) {
            int cnt;
            cin >> cnt;
            while (cnt--) {
                int x;
                cin >> x;
                treeB[i].push_back(x);
            }
        }

        if (m == 0 || n == 0) {
            cout << "NO" << endl;
            continue;
        }
        if (subTree(treeA, treeB, valA, valB, 1)) {
            cout << "YES" << endl;
        }
        else {
            cout << "NO" << endl;
        }
    }
}