Berland has n cities connected by
m bidirectional roads. No road connects a city to itself, and each pair of cities is connected by no more than one road. It isnot guaranteed that you can get from any city to any other one, using only the
existing roads.

The President of Berland decided to make changes to the road system and instructed the Ministry of Transport to make this reform. Now, each road should be unidirectional (only lead from one city to another).

In order not to cause great resentment among residents, the reform needs to be conducted so that there can be as few separate cities as possible. A city is consideredseparate, if no road leads into it, while it is allowed
to have roads leading from this city.

Help the Ministry of Transport to find the minimum possible number of separate cities after the reform.

The first line of the input contains two positive integers,
n and m — the number of the cities and the number of roads in Berland (2 ≤ n ≤ 100 000,1 ≤ m ≤ 100 000).

Next m lines contain the descriptions of the roads: thei-th road is determined by two distinct integersx_i, y_i
(1 ≤ x_i, y_i ≤ n,x_i ≠ y_i),
wherex_i andy_i are the numbers of the cities connected by thei-th
road.

It is guaranteed that there is no more than one road between each pair of cities, but it is not guaranteed that from any city you can get to any other one, using only roads.

Print a single integer — the minimum number of separated cities after the reform.

In the first sample the following road orientation is allowed: ,,.

The second sample: ,,,,.

The third sample: ,,,,.

题意：给出一些点和边。要求你把边变成有向边使得入度为0的点最少

分析：对于一些相连的点来说，假设形成的图中有环。那么我们一定可以从环上的某点出发使入度为0的点没有，

假设无环，那么仅仅须要一个入度为0的点就能使其它点入度不为0，那么问题就转换为推断图中是否有环了，

我们从任一个点遍历整个图，然后判环调整答案就可以

#include<cstring>

#include<string>

#include<iostream>

#include<queue>

#include<cstdio>

#include<algorithm>

#include<map>

#include<cstdlib>

#include<cmath>

#include<vector>

//#pragma comment(linker, "/STACK:1024000000,1024000000");

using namespace std;

#define INF 0x3f3f3f3f

vector<int>v[100006];

int vis[100006];

int ans;

int flag;

void dfs(int s,int f,int pre)

{

    vis[s]=1;

    for(int i=0; i<v[s].size(); i++)

    {

        if(vis[v[s][i]]&&pre!=v[s][i]) flag=-1;

        else if(!vis[v[s][i]])

        {

            dfs(v[s][i],f,s);

        }

    }

}

int main()

{

    int n,m;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        memset(vis,0,sizeof vis);

        for(int i=1; i<=n; i++) v[i].clear();

        for(int i=0; i<m; i++)

        {

            int a,b;

            scanf("%d%d",&a,&b);

            v[a].push_back(b);

            v[b].push_back(a);

        }

        ans=0;

        for(int i=1; i<=n; i++)

        {

            if(!vis[i])

            {

                flag=0;

                ans++;

                dfs(i,i,i);

                ans+=flag;

            }

        }

        printf("%d\n",ans);

    }

    return 0;

}