codeforces 659E New Reform - dblank

codeforces 659E New Reform

E. New Reform
time limit per test

1 second

#include<iostream>  
#include<cstdio>  
#include<algorithm>  
#include<map>  
#include<cstring>  
#include<map>  
#include<set>  
#include<queue>  
#include<stack>  
#include<cmath>  
using namespace std;  
const double eps = 1e-10;  
const int inf = 0x3f3f3f3f, N = 1e5 +10, Mod = 1000000007;  
typedef long long ll;  
int x[N], y[N], root[N], num[N], deg[N];  
int Find(int x)  
{  
    return x == root[x] ? x :root[x] = Find(root[x]);  
}  
int main()  
{  
    int n, m;  
    while(cin>>n>>m)  
    {  
        int ans = 0;  
        for(int i = 1; i<=n; i++) root[i] = i, num[i] = 1, deg[i] = 0;  
        for(int i = 0; i<m; i++)  
        {  
            scanf("%d%d", &x[i], &y[i]);  
            int u = Find(x[i]), v = Find(y[i]);  
            if(u == v) deg[u] ++;  
            else  
            {  
                root[u] = v;  
                deg[v] += deg[u] + 1;  
                num[v] += num[u];  
                deg[u] = num[u] = 0;  
            }  
        }  
        for(int i = 1; i<=n; i++) if(deg[i] < num[i]) ans++;  
        cout<<ans<<endl;  
    }  
    return 0;  
}  
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 0001 ≤ 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 ≤ nxi ≠ 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
4 3
2 1
1 3
4 3
output
1
input
5 5
2 1
1 3
2 3
2 5
4 3
output
0
input
6 5
1 2
2 3
4 5
4 6
5 6
output
1

Note

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

The second sample: .

The third sample: .

题意:给你一些边,起初这些边是无向的,最后让你把这些边变成有向的后,最小的入度为0的点最少是多少。
思路:对于有x-1条边的x个点的集合,我们可以最少得到1个入度为0的点,如果这个集合的边超过了x-1,我们可以把那一个入度为0的点消除,这样这个集合的入度为0的点就没有了,所以我们只要统计有多少集合的边等于x-1就好了,可以用并查集。

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