This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <iostream>
#include <fstream>
#include <vector>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <list>
#include <utility>
#include <cmath>
#include <numeric>
#include <assert.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pb push_back
#define endl "\n"
#define all(x) x.begin(), x.end()
const int M = 200007;
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const double pi = acos(-1);
const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
set < pair <string,int> > s; int tot;
int compress(string x)
{
auto it = s.lower_bound(make_pair(x, 0));
if(it == s.end() || (*it).F != x){
s.insert({x, ++tot});
return tot;
}
return (*it).S;
}
bool vist[M], oncycle[M], done[M];
int get(vector <int> v)
{
int n = v.size();
int ret = 0;
for(auto i : v) ret |= i;
if(!ret) return n / 2;
ret = 0;
for(int i = 0; i < n; ++i){
int j = (i + 1) % n;
if(v[i] && !v[j]){
int len = 0;
while(!v[j]){
++len;
j = (j + 1) % n;
}
ret += len / 2;
}
}
return ret;
}
vector <int> adj[M];
int n, p[M], ans, mx[M];
void calc(int node)
{
vist[node] = 1;
for(auto i : adj[node]) if(!oncycle[i]){
calc(i);
mx[node] += mx[i];
if(!done[i]) done[node] = 1;
}
if(done[node]) ++mx[node];
return;
}
int main()
{
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
cin >> n;
if(n % 2) return cout << -1 << endl, 0;
for(int i = 1; i <= n; ++i){
string x, y;
cin >> x >> y;
int a = compress(x);
int b = compress(y);
p[a] = b;
adj[b].pb(a);
}
for(int i = 1; i <= n; ++i) if(!vist[i]){
int cur = i;
while(!vist[cur]){
vist[cur] = 1;
cur = p[cur];
}
vector <int> cycle;
int st = cur;
while(1){
cycle.pb(cur);
oncycle[cur] = 1;
cur = p[cur];
if(cur == st) break;
}
vector <int> v;
int ret = 0;
for(auto j : cycle){
calc(j);
ret += mx[j];
v.pb(done[j]);
}
if(cycle.size() == 2) ret = ret - v[0] - v[1] + 2;
else ret += get(v);
ans += ret;
}
cout << n - ans << endl;
return 0;
}
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