链接
solution
这道题简化之后其实就是找图中的环和线,我们先进行并查集操作,把能连接成环的都连接成环,剩下的不能连接的放那,然后用BFS搜索一下连通块的个数,用连通块的个数减去已经成环的个数就是剩下没有成环的个数。怎么判断成没成环,在进行并查集的时候,如果我们发现他们俩的父节点是同一个,那么就确定他们已经连接成环了,然后ans1++。
#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <unordered_map>
#include <vector>
using namespace std;
#define x first
#define y second
#define endl '\n'
#define IOS \
ios_base::sync_with_stdio(0); \
cin.tie(0); \
cout.tie(0);
typedef long long ll;
typedef pair<int, int> PII;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f, mod = 1000000007;
const int N = 2e5 + 10;
int p[N];
vector<int> size(N, 1), g[N];
int find(int x)
{
if (x != p[x])
p[x] = find(p[x]);
return p[x];
}
map<int, int> vis;
void dfs(int x)
{
vis[x] = 1;
for (auto i : g[x]) {
if (!vis[i]) {
dfs(i);
}
}
}
void solve()
{
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; i++)
p[i] = i;
int ans1 = 0, ans2 = 0;
vector<PII> q;
while (m--) {
int a, c;
char b, d;
cin >> a >> b >> c >> d;
g[a].push_back(c);
g[c].push_back(a);
int fa = find(a), fc = find(c);
if (fa == fc) {
ans1++;
} else {
p[fc] = fa;
}
}
// cout << ans2 << endl;
for (int i = 1; i <= n; i++) {
if (!vis[i]) {
dfs(i);
ans2++;
}
}
cout << ans1 << " " << ans2 - ans1 << endl;
}
int main()
{
// IOS;
int t = 1;
// cin >> t;
while (t--) {
solve();
}
return 0;
}