洛谷__P1333 瑞瑞的木棍（欧拉路径）

题目链接：P1333 瑞瑞的木棍 - 洛谷

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题目大意：

给定若干木棍，每根两端有颜色。

要求判断能否将所有木棍连成一条线，使得相邻木棍接触端颜色相同。

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思路：

欧拉路径裸体，坑点是要判断图是否联通 (并查集)，若不连通肯定是不能连成一条线的

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代码：

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>//pb_ds库
#include<cctype>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second
#define  mst(a,x) memset(a,x,sizeof (a))
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef pair<int, int> pii;

const int N = 500086, mod = 998244353;

int n, m;
int d[N];
int p[N];
gp_hash_table<string, int> mp;
int idx = 0;

int find(int u) {
    if (p[u] == u) return u;
    return p[u] = find(p[u]);
}

void solve() {

    int t = 0;
    string a, b;
    for (int i = 1; i < N; i++) p[i] = i;
    while (cin >> a >> b) {
        if (!mp[a]) mp[a] = ++idx;
        if (!mp[b]) mp[b] = ++idx;
        
        int t1 = mp[a], t2 = mp[b];
        d[t1]++, d[t2]++;
        t1 = find(t1), t2 = find(t2);
        if (t1 != t2) p[t1] = t2;
        t = t2;
    }
    
    for (int i = 1; i <= idx; i++) {
        if (find(i) != find(t)) {//不连通肯定不能连成一条线
            cout << "Impossible" << endl;
            return;
        }
    }
    
    int cnt = 0;
    for (int i = 1; i <= idx; i++) {
        if (d[i] & 1) cnt++;
    }
    //欧拉路径判断
    if (cnt == 0 || cnt == 2) cout << "Possible" << endl;
    else cout << "Impossible" << endl;
    
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
// cin >> T;
    while (T--) solve();
    
    return 0;
}