Codeforces Round #346 (Div. 2) E题 并查集找环

E. New Reform

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.

Input

Output

题意：n个城市，m条双向路，将这些路改成单向的，如果一个城市没有通向它的路，（入度为0）就说明该城市是单独的。问修改后最少有几个单独的城市，要使结果最小。

思路：

1.可以建成一个有向图，可能有k个联通块，如果一个联通块没有环，就说明这个联通块，至少有一个城市单独的，因此就化成找联通块和环的问题
2.联通块的话可以用并查集来维护，然后用cir[maxn]数组来标记是否有环，如果这个联通块的根节点存在环，那么该联通块不存在单独的城市，如果不存在环的话cnt++，最后的cnt就是答案；

AC代码：

#include<bits/stdc++.h>

using namespace std;
#define N 150000
int f[N];
int arr[N];
int getf(int v){
    if(v==f[v]){
        return f[v];
    }
    f[v]=getf(f[v]);
    return f[v];
}
void merge(int u,int v){
    int t1=getf(u);
    int t2=getf(v);
    if(t1!=t2){
        f[t2]=t1;
        if(arr[t2]){
            arr[t1]=t2;
        }
    }else{
        arr[t1]=;
    }
}
int n,m;
void init(){
    ;i<=n;i++)
        f[i]=i;
}
int main(){

    cin>>n>>m;
    init();
    ;
    ;i<=m;i++){
        int x,y;
        cin>>x>>y;
        merge(x,y);
    }
    ;i<=n;i++){
        if(f[i]==i&&!arr[i]){
            ans++;
        }
    }
    cout<<ans;
    ;
}