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.
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.
Print out the only integer — the number of good paths in Uzhlyandia.
5 4
1 2
1 3
1 4
1 5
6
5 3
1 2
2 3
4 5
0
2 2
1 1
1 2
1
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.
题意：n个点 m条路 问你有多少条好路（有自环的情况）
好路就是走过所有点 m-2条路走2遍 剩下两条路走一遍
我开始有点懵 后来发现只要是联通的 那么就可以满足这个条件 问题是选两条路走一遍
自环t1 其他 t2
1》两条都不是自环的 那么这两条一定有公共点
for(int i=1;i<=n;i++)ans=ans+a[i]*(a[i]-1)/2;
2》一条有自环 一条不是 t1*t2;
3》都是自环 t1*(t1-1)/2
开始edge[N]开小了 无限wa 懵比了
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<map> 5 #include<cstdlib> 6 #include<vector> 7 #include<set> 8 #include<queue> 9 #include<cstring> 10 #include<string.h> 11 #include<algorithm> 12 typedef long long ll; 13 typedef unsigned long long LL; 14 using namespace std; 15 const int N=1000000+100; 16 int m,n; 17 int flag[N]; 18 int head[N]; 19 int vis[N]; 20 ll a[N],b[N]; 21 int cnt=0; 22 void init(){ 23 memset(vis,0,sizeof(vis)); 24 cnt=0; 25 memset(head,-1,sizeof(head)); 26 } 27 struct node{ 28 int to,next; 29 }edge[N*2+100]; 30 void add(int u,int v){ 31 edge[cnt].to=v; 32 edge[cnt].next=head[u]; 33 head[u]=cnt++; 34 } 35 void DFS(int x){ 36 vis[x]=1; 37 for(int i=head[x];i!=-1;i=edge[i].next){ 38 int v=edge[i].to; 39 if(vis[v]==0){ 40 DFS(v); 41 } 42 } 43 } 44 int main(){ 45 scanf("%d%d",&n,&m); 46 init(); 47 ll t1=0; 48 ll t2=0; 49 int u,v; 50 memset(a,0,sizeof(a)); 51 for(int i=0;i<m;i++){ 52 scanf("%d%d",&u,&v); 53 b[u]=b[v]=1; 54 if(u!=v){add(u,v);add(v,u);t1++;a[u]++;a[v]++;} 55 else 56 t2++; 57 } 58 DFS(u); 59 for(int i=1;i<=n;i++){ 60 if(vis[i]==0&&b[i]){ 61 cout<<0<<endl; 62 return 0; 63 } 64 } 65 ll ans=0; 66 ans=ans+t1*t2; 67 ans=ans+(t2*(t2-1))/2; 68 for(int i=1;i<=n;i++) 69 ans=ans+(a[i]*(a[i]-1))/2; 70 printf("%I64d ",ans); 71 }