codeforces 789 D. Weird journey （欧拉路计数）

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题意：给你一张n个点m条边的图，问你经过其中m-2条边两次，剩下2条边一次的方案数有几种，如果剩下两条边的集合一样算同一种

题解：把题意转换成，我们挑选两条边，把其他的边都变成两条，这样就变成了一笔画问题(欧拉路径是否存在)，我们欧拉路径存在的充要条件是奇点个数是0或2,。然后下面需要进行分类讨论，我们把边分成普通变和自环（u=v）两类。

1.选取两条不相邻普通边，图中存在4个奇点，不满足欧拉路径条件

2.选取两条相邻普通边，图中存在2个奇点，满足欧拉路径条件

3.选取一条普通边一条自环，图中存在2个奇点，满足欧拉路径条件

4.选取两条自环，图中存在0个奇点，满足欧拉路径条件

在这之前如果m条边覆盖的集合不是连通的，答案为0

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 41 21 31 41 5
output
6
input
5 31 22 34 5
output
0
input
2 21 11 2
output
1

Note

In first sample test case the good paths are:

2 → 1 → 3 → 1 → 4 → 1 → 5,
2 → 1 → 3 → 1 → 5 → 1 → 4,
2 → 1 → 4 → 1 → 5 → 1 → 3,
3 → 1 → 2 → 1 → 4 → 1 → 5,
3 → 1 → 2 → 1 → 5 → 1 → 4,
4 → 1 → 2 → 1 → 3 → 1 → 5.

There are good paths that are same with displayed above, because the sets of roads they pass over once are same:

2 → 1 → 4 → 1 → 3 → 1 → 5,
2 → 1 → 5 → 1 → 3 → 1 → 4,
2 → 1 → 5 → 1 → 4 → 1 → 3,
3 → 1 → 4 → 1 → 2 → 1 → 5,
3 → 1 → 5 → 1 → 2 → 1 → 4,
4 → 1 → 3 → 1 → 2 → 1 → 5,
and all the paths in the other direction.

Thus, the answer is 6.

In the second test case, Igor simply can not walk by all the roads.

In the third case, Igor walks once over every road.

#include<bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6 + 100;typedef long long ll;vector<int>vec[N];int a[N],fa[N],vis[N];int deg[N];int find(int x){    int r=x;    while(fa[r]!=r) r=fa[r];    int i=x,j;    while(i!=r) {        j=fa[i];        fa[i]=r;        i=j;    }    return r;}int main(){    ios::sync_with_stdio(false);    int n,m,i,j,u,v;    int fx,fy;    ll loop,ans,edge;    cin>>n>>m;    loop=ans=edge=0;    for(i=1;i<=n;i++) fa[i]=i;    while(m--) {        cin>>u>>v;        vis[u]=vis[v]=1;        if(u==v) {            loop++;            continue;        }        edge++;        vec[u].push_back(v);        vec[v].push_back(u);        deg[u]++;        deg[v]++;        fx=find(u);        fy=find(v);        if(fx!=fy) fa[fx]=fy;    }    for(j=1;j<=n;j++)        if(vis[j]) break;    for(i=1;i<=n;i++)        if(vis[i] && find(i)!=find(j)) break;    if(i<=n) {        printf("0\n");    }    else {        for(i=1;i<=n;i++) {            ans+=(ll) deg[i]*(deg[i]-1)/2;        }        ans+=(ll)loop*edge;        ans+=(ll) loop*(loop-1)/2;        cout<<ans<<endl;    }    return 0;}

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