Codeforce Div-3 E.Cyclic Components

You are given an undirected graph consisting of $\mathrm{nn vertices\; and mm edges.\; Your\; task\; is\; to\; find\; the\; number\; of\; connected\; components\; which\; are\; cycles.}$

Here are some definitions of graph theory.

An undirected graph consists of two sets: set of nodes (called vertices) and set of edges. Each edge connects a pair of vertices. All edges are bidirectional (i.e. if a vertex $\mathrm{aa is\; connected\; with\; a\; vertex bb,\; a\; vertex bb is\; also\; connected\; with\; a\; vertex aa).\; An\; edge\; can\text{'}t\; connect\; vertex\; with\; itself,\; there\; is\; at\; most\; one\; edge\; between\; a\; pair\; of\; vertices.}$

Two vertices $\mathrm{uu and vv belong\; to\; the\; same\; connected\; component\; if\; and\; only\; if\; there\; is\; at\; least\; one\; path\; along\; edges\; connecting uu and vv.}$

A connected component is a cycle if and only if its vertices can be reordered in such a way that:

• the first vertex is connected with the second vertex by an edge,
• the second vertex is connected with the third vertex by an edge,
• ...
• the last vertex is connected with the first vertex by an edge,
• all the described edges of a cycle are distinct.

A cycle doesn't contain any other edges except described above. By definition any cycle contains three or more vertices.

There are $\mathrm{66 connected\; components, 22 of\; them\; are\; cycles: [7,10,16][7,10,16] and [5,11,9,15][5,11,9,15].}$

Input

The first line contains two integer numbers $\mathrm{nn and mm (1\le n\le 2\cdot 1051\le n\le 2\cdot 105, 0\le m\le 2\cdot 1050\le m\le 2\cdot 105)\; —\; number\; of\; vertices\; and\; edges.}$

The following $\mathrm{mm lines\; contains\; edges:\; edge ii is\; given\; as\; a\; pair\; of\; vertices vivi, uiui (1\le vi,ui\le n1\le vi,ui\le n, ui\ne viui\ne vi).\; There\; is\; no\; multiple\; edges\; in\; the\; given\; graph,\; i.e.\; for\; each\; pair\; (vi,uivi,ui)\; there\; no\; other\; pairs\; (vi,uivi,ui)\; and\; (ui,viui,vi)\; in\; the\; list\; of\; edges.}$

Output

Print one integer — the number of connected components which are also cycles.

Examples

input

5 4
1 2
3 4
5 4
3 5

output

1

input

17 15
1 8
1 12
5 11
11 9
9 15
15 5
4 13
3 13
4 3
10 16
7 10
16 7
14 3
14 4
17 6

output

2

Note

In the first example only component $[3,4,5][3,4,5] is\; also\; a\; cycle.$

The illustration above corresponds to the second example.

 

题意:

　　让你求回路的个数,而且这个回路是没有杂边的单环。

思路

　　如果有这样的回路，那么每一个节点的度数一定为2。用dfs跑一遍，如果在跑的过程中，所有的点度数均为2，那么它一定就是我们要找的环。

　　这里我使用了一种新的邻接表，使用vector，这种方式没有办法存储边的长度，但是可以很直接的看出点的度数。如果要用vector来储存权值的话，那么再开一个vector，依次记录就行了

#include<iostream>
#include<vector>
using namespace std;
int book[200005];
int g;
vector<int>a[200005];

void dfs(int x)
{
    book[x]=1;
    if(a[x].size()!=2){g=1;}

    for(int i:a[x]){
        if(!book[i]){dfs(i);}
    }
}

int main()
{
    int n,m;
    cin>>n>>m;
    int x,y;
    for(int i=1;i<=m;i++){
        cin>>x>>y;
        a[x].push_back(y);
        a[y].push_back(x);
    }
    int ans=0;
    for(int i=1;i<=n;i++){
        g=0;
        if(!book[i]){
            dfs(i);
            if(!g){ans++;}
        }
    }
    cout<<ans<<endl;
}

以上思路来自于大神代码：