若能一筆畫,則滿足
1.所有節點皆為偶數邊 (一進一出)
2.若有奇數邊,必剛好有兩組 (一個起點,一個終點)
#include <iostream>
#include <map>
using namespace std;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int vertex, edge;
int v1, v2, odd;
map<int, int> connect_way;
while(cin >> vertex >> edge)
{
connect_way.erase(connect_way.begin(), connect_way.end());
for(int i = 0; i<edge; i++)
{
cin >> v1 >> v2;
connect_way[v1]++;
connect_way[v2]++;
}
odd = 0;
for(map<int,int>::iterator iter = connect_way.begin(); iter != connect_way.end(); iter++)
if(iter->second % 2)
odd++;
if(odd > 0 && odd != 2)
cout << "NO" << endl;
else
cout << "YES" << endl;
}
}