给定两个无向图
和 ，判断它们是否同构。图的同构是指两个图的节点可以通过某种重新编号的方式完全匹配，且边的连接关系一致。
为了简化问题，假设图的节点编号从 0 到 n-1，并且图的边以邻接表的形式给出。下面程序中横线处应该给出的是（ ）
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
string graphHash(vector<vector<int>>& graph) {
vector<string> nodeHashes(graph.size());
for (int i = 0; i < graph.size(); i++) {
vector<int> neighbors = graph[i];
sort(neighbors.begin(), neighbors.end());
string hash;
for (int neighbor : neighbors) {
____________
}
nodeHashes[i] = hash;
}
sort(nodeHashes.begin(), nodeHashes.end());
string finalHash;
for (string h : nodeHashes) {
finalHash += h + ";";
}
return finalHash;
}
int main() {
int n;
cin >> n;
vector<vector<int>> G1(n);
for (int i = 0; i < n; i++) {
int k;
while (cin >> k) {
G1[i].push_back(k);
if (cin.get() == '\n') break;
}
}
vector<vector<int>> G2(n);
for (int i = 0; i < n; i++) {
int k;
while (cin >> k) {
G2[i].push_back(k);
if (cin.get() == '\n') break;
}
}
string hash1 = graphHash(G1);
string hash2 = graphHash(G2);
if (hash1 == hash2) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
- A.
hash += to_string(neighbor); - B.
hash += to_string(neighbors); - C.
hash += to_string(neighbor) + ","; - D.
hash -= to_string(neighbors);