題目看著好嚇人好嚇人。

其實呢。。

就是統計度的個數和圖是否聯通。。。。。

#include <iostream>#include <vector>#include <cstring>#include <algorithm>#define MAX 510using namespace std;vector <int> g[MAX];int v, e;bool isVis[MAX];void DFS(int st) {	isVis[st] = true;	for (int i = 0; i < g[st].size(); i++) {		if(!isVis[g[st][i]])			DFS(g[st][i]);	}}bool DFSTrave() {	int count = 0;	for (int i = 1; i <= v; i++) {		if (!isVis[i])			DFS(i),count++;	}	return count == 1;}int main() {	int x, y;	cin >> v >> e;	for (int i = 0; i < e; i++) {		cin >> x >> y;		g[x].push_back(y);		g[y].push_back(x);	}	memset(isVis, false, sizeof(isVis));	int odd = 0, even = 0;	for (int i = 1; i < v; i++) {		if (g[i].size() % 2 == 0) {			even++;		}		else {			odd++;		}		cout << g[i].size() << " ";	}	if (g[v].size() % 2 == 0)		even++;	else		odd++;	cout << g[v].size() << endl;	bool tag = DFSTrave();	if (odd == 0 && tag)		cout << "Eulerian" << endl;	else if (odd == 2 && tag)		cout << "Semi-Eulerian" << endl;	else		cout << "Non-Eulerian" << endl;		return 0;}