lightoj1300

思路：显然是和边双连通分量有关的，所以只需要在双连通分量中找奇环（二分图染色），如果找到，这个连通分量中的点都是满足的。

// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
// #define DEBUG
#ifdef DEBUG
#define debug(...) printf( __VA_ARGS__ )
#else
#define debug(...)
#endif
#define MEM(x,y) memset(x, y,sizeof x)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 10010;
int dfn[maxn], low[maxn], Belong[maxn], num[maxn];
int Time;
int BCC;
int n, m;
vector<int> G[maxn];
void Init(){
	for (int i = 0;i < n;++i)
		G[i].clear();
	Time = BCC = 0;
	memset(dfn, -1,sizeof dfn);
}
void Read(){
	int u, v;
	for (int i = 0;i < m;++i){
		scanf("%d%d",&u,&v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
}
stack<int> st;
void Tarjan(int u,int fa){
	dfn[u] = low[u] = Time++;
	st.push(u);
	int cnt = 0;
	for (int i = 0;i < G[u].size();++i){
		int v = G[u][i];
		if (v == fa && cnt == 0){
			cnt = 1;
			continue;
		}
		if (dfn[v] == -1){
			Tarjan(v, u);
			low[u] = min(low[u], low[v]);
		}else low[u] = min(low[u], dfn[v]);
	}
	if (low[u] == dfn[u]){
		BCC++;
		num[BCC] = 0;
		while(true){
			int v = st.top();
			st.pop();
			num[BCC]++;
			Belong[v] = BCC;
			if (u == v) break;
		}
	}
}
int Color[maxn];
bool vis[maxn];
bool DFS(int u, int ID){
	for (int i = 0;i < G[u].size();++i){
		int v = G[u][i];
		if (Belong[v] != ID) continue;
		if (Color[u] == Color[v]) return false;//奇环
		if (Color[u] + Color[v] == 3) continue;//偶环
		if (Color[v] == -1){
			Color[v] = 3 - Color[u];
			if (!DFS(v, ID)) return false;
		}
	}
	return true;
}
int icase = 0;
void solve(){
	for (int i = 0;i < n;++i)
		if (dfn[i] == -1) Tarjan(i, -1);
	// printf("BCC = %d\n", BCC);
	memset(Color, -1,sizeof Color);
	memset(vis, false,sizeof vis);
	int sum = 0;
	for (int i = 0;i < n;++i){
		int ID = Belong[i];
		if (vis[ID]) continue;
		vis[ID] = true;
		Color[i] = 1;
		if (!DFS(i, ID)) sum += num[ID];
	}
	printf("Case %d: %d\n", ++icase, sum);
}
int main()
{	
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	int t;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&m);
		Init();
		Read();
		solve();
	}
	return 0;
}