Codeforces Round #407 (Div. 1) B. Weird journey

RJ28發表於2017-04-02

Little boy Igor wants to become a traveller. At first, he decided to visit all the cities of his motherland — Uzhlyandia.

It is widely known that Uzhlyandia has n cities connected with m bidirectional roads. Also, there are no two roads in the country that connect the same pair of cities, but roads starting and ending in the same city can exist. Igor wants to plan his journey beforehand. Boy thinks a path is good if the path goes over m - 2 roads twice, and over the other 2 exactly once. The good path can start and finish in any city of Uzhlyandia.

Now he wants to know how many different good paths are in Uzhlyandia. Two paths are considered different if the sets of roads the paths goes over exactly once differ. Help Igor — calculate the number of good paths.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 106) — the number of cities and roads in Uzhlyandia, respectively.

Each of the next m lines contains two integers u and v (1 ≤ u, v ≤ n) that mean that there is road between cities u and v.

It is guaranteed that no road will be given in the input twice. That also means that for every city there is no more than one road that connects the city to itself.

Output

Print out the only integer — the number of good paths in Uzhlyandia.

Examples
input
5 4
1 2
1 3
1 4
1 5
output
6
input
5 3
1 2
2 3
4 5
output
0
input
2 2
1 1
1 2
output

1


分析：從一個點出發如果經過的所有邊都走了兩邊,那麼相當於最後回到了起點,簡單畫一下圖就會發現符合條件的邊對要麼共頂點,要麼有一條邊是自環.


#include <bits/stdc++.h>
#define N 1000005
#define INF 2147483647
using namespace std;
int n,m,cnt,u,v,root,vis[N],f[N],edg[N];
long long ans;
vector<int> G[N];
void dfs(int u)
{
	vis[u] = true;
	for(int i = 0;i < G[u].size();i++)
	{
		int v = G[u][i];
		if(!vis[v]) dfs(v); 
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i = 1;i <= m;i++)
	{
		scanf("%d%d",&u,&v);
		edg[u] = edg[v] = true;
		if(u == v)
		{
			cnt++;
			continue;
		}
		G[u].push_back(v);
		G[v].push_back(u);
		f[u]++,f[v]++;
		root = u;
	}
	dfs(u);
	bool flag = true;
	for(int i = 1;i <= n;i++) 
	 if(edg[i]) flag &= vis[i];
	for(int i = 1;i <= n;i++) ans += 1ll*f[i]*(f[i]-1)/2;
	cout<<flag*(ans+1ll*cnt*(m-cnt)+1ll*cnt*(cnt-1)/2)<<endl; 
}


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