time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
After Vitaly was expelled from the university, he became interested in the graph theory.
Vitaly especially liked the cycles of an odd length in which each vertex occurs at most once.
Vitaly was wondering how to solve the following problem. You are given an undirected graph consisting of n vertices and m edges, not necessarily connected, without parallel edges and loops. You need to find t — the minimum number of edges that must be added to the given graph in order to form a simple cycle of an odd length, consisting of more than one vertex. Moreover, he must find w — the number of ways to add t edges in order to form a cycle of an odd length (consisting of more than one vertex). It is prohibited to add loops or parallel edges.
Two ways to add edges to the graph are considered equal if they have the same sets of added edges.
Since Vitaly does not study at the university, he asked you to help him with this task.
Input
The first line of the input contains two integers n and m ( — the number of vertices in the graph and the number of edges in the graph.
Next m lines contain the descriptions of the edges of the graph, one edge per line. Each edge is given by a pair of integers ai, bi (1 ≤ ai, bi ≤ n) — the vertices that are connected by the i-th edge. All numbers in the lines are separated by a single space.
It is guaranteed that the given graph doesn’t contain any loops and parallel edges. The graph isn’t necessarily connected.
Output
Print in the first line of the output two space-separated integers t and w — the minimum number of edges that should be added to the graph to form a simple cycle of an odd length consisting of more than one vertex where each vertex occurs at most once, and the number of ways to do this.
Examples
input
4 4
1 2
1 3
4 2
4 3
output
1 2
input
3 3
1 2
2 3
3 1
output
0 1
input
3 0
output
3 1
Note
The simple cycle is a cycle that doesn’t contain any vertex twice.
【题目链接】:http://codeforces.com/contest/557/problem/D
【题解】
要存在一个奇数环。
则最多就添加3条边(3条边一定能构成一个环!)。
1.看看整张图变成了几个连通块,如果每个连通块里面的点的个数都为1，则添加边数为3,方案数为C(n,3)=n*(n-1)*(n-2)/6,这个时候对应的情况是边数m=0;->”3 C(N,3)”
2.每个连通块里面的点的个数的最大值为2;则连通块里面点的个数为2的情况就对应这个连通块里面只有一条边,而一条边由两个点构成，这条边上的两个点分别与其余n-2个点构成n-2个环(都是3个点的环),边的个数m就对应了连通块里面点的个数为2的情况，则方案为m*(n-2);->”2 m*(n-2)”
下面这种情况不是奇环(而是偶环)，所以”2对2的情况可以排除”;
3.除了以上两种情况外;
如果在某个连通块里面不能进行二分图染色->则存在奇环。直接输出”0 1”;
( 有奇环就不能完成二分图染色);
如果都能进行二分图染色;
则记录每个连通块里面白点(0)和黑点(1)的个数;
设为cnt[0]和cnt[1];
则每有一个联通块；
答案递增C(cnt[0],2)+C(cnt[1],2);
可以看到每两个0之间连一条边都能构成一个奇数环;
很有趣的性质.
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
#define pri(x) printf("%d",x)
#define prl(x) printf("%I64d",x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int MAXN = 1e5+10;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
int n,m;
int f[MAXN],num[MAXN],cnt[2];
int color[MAXN];
vector <int> g[MAXN];
queue <int> dl;
int ff(int x)
{
if (f[x]==x) return x;
else
f[x] = ff(f[x]);
return f[x];
}
int main()
{
//freopen("F:\rush.txt","r",stdin);
memset(color,255,sizeof color);
rei(n);rei(m);
rep1(i,1,n)
f[i] = i,num[i] = 1;
rep1(i,1,m)
{
int x,y;
rei(x);rei(y);
g[x].pb(y);
g[y].pb(x);
int r1 = ff(x),r2 = ff(y);
if (r1!=r2)
{
f[r1]=r2;
num[r2]+=num[r1];
}
}
int ma = 1;
LL ans = 0;
rep1(i,1,n)
{
int r = ff(i);
ma = max(ma,num[r]);
}
if (ma == 1)
{
printf("3 %I64d
",1LL*n*(n-1)*(n-2)/6);
return 0;
}
else
if (ma==2)
{
printf("2 %I64d
",1LL*(n-2)*m);
return 0;
}
else
{
rep1(i,1,n)
if (color[i]==-1)
{
memset(cnt,0,sizeof cnt);
color[i] = 0;
cnt[0] = 1;
dl.push(i);
bool ok = true;
while (!dl.empty())
{
int x = dl.front();
dl.pop();
int len = g[x].size();
rep1(j,0,len-1)
{
int y = g[x][j];
if (y==x) continue;
if (color[y]==-1)
{
color[y] = 1-color[x];
cnt[color[y]]++;
dl.push(y);
}
else
if (color[y]==color[x])
{
ok = false;
break;
}
}
if (!ok) break;
}
if (!ok)
{
printf("0 1
");
return 0;
}
if (cnt[0]>=2)
ans+=1LL*cnt[0]*(cnt[0]-1)/2;
if (cnt[1]>=2)
ans+=1LL*cnt[1]*(cnt[1]-1)/2;
}
}
cout <<"1 "<< ans << endl;
return 0;
}