FZU2112 Tickets（欧拉通路）

You have won a collection of tickets on luxury cruisers. Each ticket can be used only once, but can be used in either direction between the 2 different cities printed on the ticket. Your prize gives you free airfare to any city to start your cruising, and free airfare back home from wherever you finish your cruising.

You love to sail and don't want to waste any of your free tickets. How many additional tickets would you have to buy so that your cruise can use all of your tickets?

Now giving the free tickets you have won. Please compute the smallest number of additional tickets that can be purchased to allow you to use all of your free tickets.

 

Input

 

There is one integer T (T≤100) in the first line of the input.

Then T cases, for any case, the first line contains 2 integers n, m (1≤n, m≤100,000). n indicates the identifier of the cities are between 1 and n, inclusive. m indicates the tickets you have won.

Then following m lines, each line contains two integers u and v (1≤u, v≤n), indicates the 2 cities printed on your tickets, respectively.

 

 

 

Output

For each test case, output an integer in a single line, indicates the smallest number of additional tickets you need to buy.

 

Sample Input

 

3
5 3
1 3
1 2
4 5
6 5
1 3
1 2
1 6
1 5
1 4
3 2
1 2
1 2

Sample Output

 

 

1
2
0

题解：

 

每次得到一条边时，把两个城市的度都加一，最后得出有多少个度为奇数的城市，因为第一个城市和最后一个城市有接送所以减去二，剩下的城市数除以二得到的就是结果。注意当没有度为奇数的城市时直接输出0。（总结一句话，每个城市有进就地有出，而票都是双向的所以只要度为偶数就满足。）

代码：

 

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

int board[100005];

int main()
{
	int N;
	cin>>N;
	while(N--)
	{
		int n,m;
		memset(board,0,sizeof(board));
		scanf("%d %d",&n,&m);
		while(m--)
		{
			int a,b;
			scanf("%d %d",&a,&b);
			board[a]++;
			board[b]++;
		}
		int sum = 0;
		for(int i=1 ; i<=n ; i++)
		{
			if(board[i]&1)sum++;
		}
		if(sum == 0);
		else
		{
			sum -= 2;
			if(sum)sum /= 2;
		}
		printf("%d\n",sum);
	}

	return 0;
}