题意：
思路：
种类并查集， 具体见代码
注意判断矛盾的方法：如果两个节点的属于同一个集合 判断的是它们而不是它们的根节点（代码68行）
最后统计个数的方法需要更加熟练！
// Decline is inevitable,
// Romance will last forever.
#include <bits/stdc++.h>
using namespace std;
//#define mp make_pair
#define pii pair<int,int>
#define pb push_back
#define fi first
#define se second
#define ll long long
#define LL long long
//#define int long long
const int maxn = 3e5 + 10;
const int maxm = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int dx[] = {0, 0, -1, 1}; //{0, 0, 1, 1, 1,-1,-1,-1}
const int dy[] = {1, -1, 0, 0}; //{1,-1, 1, 0,-1, 1, 0,-1}
const int P = 1e9 + 7;
int n, m;
int fa[maxn];
int vis[maxn];
int cnt[maxn];
int cnt2[maxn];
map<int, int> mp;
struct node {
int fa;
int re; //0表示和根节点同类 1表示反类
}a[maxn];
int ans;
int find(int x){ //查找x的祖先
if(x == a[x].fa) return x;
int tmp = a[x].fa;
a[x].fa = find(a[x].fa);
a[x].re = (a[x].re + a[tmp].re) % 2;
return a[x].fa;
}
void merge(int x, int y){ //把y合并到x家族
int fx = find(x);
int fy = find(y);
if(fx != fy) {
fa[fx] = fy;
}
}
void init() {
for(int i=1; i <= n; i++) {
a[i].fa = i;
a[i].re = 0;
vis[i] = cnt[i] = cnt2[i] = 0;
}
ans = 0;
}
void solve() {
mp.clear();
cin >> n >> m;
init();
bool ok = true;
while(m--) {
int u, v;
string s;
cin >> u >> v >> s;
int d = 1;
if(u == v) continue;
if(s[0] == 'c') d = 0;
int fx = find(u);
int fy = find(v);
if(fx == fy) {
if((a[u].re + a[v].re) % 2 != d)
ok = false;
}
else if(fx != fy) {
a[fy].fa = fx;
a[fy].re = (a[u].re + a[v].re + d) % 2;
}
}
if(!ok) {
cout << -1 << '\n';
return;
}
int p = 0;
for(int i = 1; i <= n; i++) {
int fx = find(i);
if(!vis[fx]) {
mp[fx] = ++p;
vis[fx] = 1;
}
}
for(int i = 1; i <= n; i++) {
int fx = find(i);
cnt[mp[fx]]++;
if(a[i].re == 1) cnt2[mp[fx]]++;
}
ans = 0;
for(int i = 1; i <= p; i++) {
ans = ans + max(cnt2[i], cnt[i] - cnt2[i]);
}
cout << ans << '\n';
}
int main() {
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
// int T; scanf("%d", &T); while(T--)
// freopen("1.txt","r",stdin);
// freopen("2.txt","w",stdout);
int T; cin >> T;while(T--)
solve();
return 0;
}
/*
*/