E. New Reform
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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 is not 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 considered separate, 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.
Input
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: the i-th road is determined by two distinct integers xi, yi (1 ≤ xi, yi ≤ n, xi ≠ yi), where xi and yi are the numbers of the cities connected by the i-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.
Output
Print a single integer — the minimum number of separated cities after the reform.
Examples
input
Copy
4 3 2 1 1 3 4 3
output
Copy
1
input
Copy
5 5 2 1 1 3 2 3 2 5 4 3
output
Copy
0
input
Copy
6 5 1 2 2 3 4 5 4 6 5 6
output
Copy
1
Note
In the first sample the following road orientation is allowed: 
, 
, 
.
The second sample: , 
, 
, 
, .
The third sample: , , 
, 
, 
.
题目大意: 给m条无向路，看怎样给方向使得n个城市中，孤立城市（没有指向它的路的城市）最少
可以假定这些都是无向路（即双向路），然后把每个图分成好多个连通块，如果每一个小连通图中有环，则孤城为0个，
如果无环，则孤城为1；
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
//每一个小的连通图如果有环，没有所谓的孤城，如果没有环，那么有一个孤城；
const int maxn = 100000+5;
vector <int> G[maxn];//用邻接表存
int v[maxn];//标记是否走过
int ans,res;
void dfs(int x,int y){//y是这个点连接的上一个点
if(v[x]){//如果访问过，证明有环的存在，res=0，返回，否则，继续找连通图，如果所有点都被标记完了，那么res不变，为初始值1
res = 0;
return;
}
v[x] = 1;//标记走过的点
for(int i = 0;i < G[x].size();i++)
if(G[x][i]!=y){//遍历x连接的所有点，除了x连接的上一个点
dfs(G[x][i],x);
}
}
int main()
{
int n,m,i,a,b;
while(cin >> n >> m){
memset(v,0,sizeof(v));
for(i = 0;i <= n;i++)
G[i].clear();
for(i = 0;i < m;i++){
cin >> a >> b;
G[a].push_back(b);//相当于无向边
G[b].push_back(a);
}
ans = 0;
for(i = 1;i <= n;i++)
if(!v[i]){
res = 1;
dfs(i,0);//初始时把所有点的上一个点当作0，dfs求连通图并标记
ans += res;
}
cout << ans << endl;
}
return 0;
}