欧拉求解

东软学习小组成员：夜枫

测试案例

#include
#include  
#include
using namespace std;
stack<int> s;
stack<int> q;
const int N = 100;
int map[N][N];//矩阵 
int ans[N];//顶点
int n, m;//n个顶点，m条边 
bool vis[N] = {false};
int set[1000005];

int find(int x){

	return x == set[x] ? x : (set[x] = find(set[x]));   //递归查找集合的代表元素，含路径压缩。

}
void dfs1(int s){             //递归深搜
	vis[s] = true;
	for (int i = 1; i<=n; ++i){
		if (!vis[i]){
			dfs1(i);
		}
	}
}

bool judge(){                //判断是否所有点已被遍历过
	for (int i = 1; i <= n; ++i)
		if (!vis[i])
			return false;
	return true;
}

void dfs(int x, int isyou = 0){//默认无向图//isyou=1为有向图 
	s.push(x);
	for (int i = 1; i <= n; ++i){
		if (map[x][i] >= 1){//有边
			map[x][i]--;
			if (isyou == 0){//无向图对称 
				map[i][x]--;
			}
			dfs(i, isyou);
			break;//一次 
		}
	}
}
void fleury(int x,int isyou){
	s.push(x);
	int t = x;
	while (!s.empty()){
		int k = 0;
		t = s.top();
		s.pop();
		for (int i = 1; i <= n; ++i){
			if (map[t][i] >= 1){//有边
				k = 1;
				break;
			}
		}
		if (k == 0){
			q.push(t);
		}
		else{
			dfs(t,isyou);
		}
	}
	while (!q.empty()){
		cout << q.top() << " ";
		q.pop();
	}
}
int main(){
	cin >> n >> m;
	int x, y,isyou;
	int du[N];
	int duc[N];
	int sum = 0;
	
	for (int i = 1; i < 1000005; ++i)        //初始化个集合，数组值等于小标的点为根节点。
	{

		set[i] = i;
	}

	memset(du, 0, sizeof(du));
	memset(vis, false, sizeof(vis));
	memset(duc, 0, sizeof(duc));
	memset(map, 0, sizeof(map));
	cout << "是否为有向图  1.是，0,不是\n";
	cin >> isyou;
	for (int i = 1; i <= m; ++i){
		cin >> x >> y;

		int fx = find(x), fy = find(y);
		set[fx] = fy;

		map[x][y]++;
		duc[x]++;
		du[y]++;
		if (isyou == 0){
			map[y][x]++;
			duc[y]++;
			du[x]++;
		}
		
	}



	int duflag;

	int flag = 1;
	for (int i = 1; i <= n; ++i){
		if ((du[i] + duc[i]) % 2 == 1){
			sum++;
			if (du[i]<duc[i]){
				flag = i;
			}
		}
	}
	cout << "sum " << sum << endl;

	if (sum == 0){

		int cnt = 0;

		for (int i = 1; i <= n; ++i)          //统计集合个数，即为连通分量个数，为一时，图联通。
		{
			if (set[i] == i)
			{
				++cnt;

			}
		}
		if (cnt != 1){
			cout << "你输入的是非欧拉图" << endl;
			exit(0);
		}



		if (isyou == 1){//无向图
			for (int i = 1; i <= n; ++i){
				if (du[i] != duc[i]){
					cout << "你输入的是非欧拉图" << endl;
					exit(0);
				}
			}
		}

		if (isyou == 1){
			for (int i = 1; i <= n; ++i){
				if ((du[i]+duc[i])%2==1){
					cout << "你输入的是非欧拉图" << endl;
					exit(0);
				}
			}
		}

	}

	if (sum == 2){
	
			int cnt = 0;

			for (int i = 1; i <= n; ++i)          //统计集合个数，即为连通分量个数，为一时，图联通。
			{
				if (set[i] == i)
				{
					++cnt;

				}
			}

			if (cnt == 1){
				int min = 0, max = 0;
				for (int i = 1; i <= n; ++i){
					if ((du[i] + duc[i]) % 2 == 1){
						if (du[i] < duc[i]){
							min = 1;
						}
						if (du[i] > duc[i]){
							max = 1;
						}
					}
				}
				if (min != 1 || max != 1){
					cout << "你输入的是非欧拉图" << endl;
					exit(0);
				}
			}
			else{

				cout << "你输入的是非欧拉图" << endl;
				exit(0);

			}

		

	}


	if (sum == 0 || sum == 2){
	//	cout << "开始" << endl;
		fleury(flag,isyou);
	}
	return 0;
}
/*
案例一
4 4
0
1 2
2 3
3 4
4 1

案例二
6 7
1
2 1
2 4
4 3
3 2
1 6
6 5
5 1

*/