题解 | 鞋带难题

#include <iostream>
#include <queue>
#include <vector>
#include <cstring>

using namespace std;

int main() {
	int n, m;
	cin >> n >> m;
	vector<vector<int> > graph(n + 1);
	int v1, v2;
	int* degree = new int[n + 1];
	memset(degree, 0, (n + 1) * sizeof(int));
	for (int i = 0; i < m; i++) {
		cin >> v1 >> v2;
		graph[v1].push_back(v2);
		graph[v2].push_back(v1); 
		degree[v1]++;
		degree[v2]++;
	}
	int* dest = new int[n + 1];
	int* vis = new int[n + 1];
	memset(dest, 0xFF, (n + 1) * sizeof(int));
	memset(vis, 0, (n + 1) * sizeof(int));
	queue<int> q;
	for (int i = 1; i < n + 1; i++) {
		if (degree[i] == 1) {
			q.push(i);
			dest[i] = 1;
		}
	}
//	while (!q.empty()) {
//		int v = q.front();
//		q.pop();
//		vis[v] = 1;
//		if (degree[v] == 0) dest[v]--;之前没有加这一步判定会导致结果可能大1
//		for (vector<int>::iterator it = graph[v].begin(); it != graph[v].end(); it++) {
//			degree[*it]--;
//			if (!vis[*it] && degree[*it] == 1) {
//				q.push(*it);
//				dest[*it] = dest[v] + 1;
//			}
//		}
//	}
//	int maxVal = 0;
//	for (int i = 1; i < n + 1; i++) {
//		maxVal = max(maxVal, dest[i]);
//	}
//	cout << maxVal << endl; 
	
  	
  	//更直观的写法
	int ans = 0;
	while (!q.empty()) {
		int sz = q.size();
		bool removed = false;
		for (int i = 0; i < sz; i++) {
			int v = q.front();
			q.pop();
			if (degree[v] == 0) continue;
			degree[v] = 0;
			removed = true;
			for (vector<int>::iterator it = graph[v].begin(); it != graph[v].end(); it++) {
				degree[*it]--;
				if (degree[*it] == 1) q.push(*it);
			}
		}
		ans += removed ? 1 : 0;
	}
	cout << ans << endl;
}