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TangDou/AcWing/EulerPath/HDU1878.md

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+## $HDU$ $1878$ 无向图判定欧拉回路[模板题] 
+![](https://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/%7Byear%7D/%7Bmonth%7D/%7Bmd5%7D.%7BextName%7D/20230808150148.png)
+
+定理：无向图 $G$ 具有一条欧拉回路，当且仅当 $G$ 是连通的，并且所有结点度数为偶数。
+
+思路：不需要建图。并查集统计无向图中连通块的个数，开一个数组统计每个点的度数。
+
+### $Code$
+```cpp {.line-numbers}
+// https://blog.51cto.com/u_3044148/5225981
+
+/*
+输入样例
+3 3
+1 2
+1 3
+2 3
+3 2
+1 2
+2 3
+0
+
+输出样例
+1
+0
+*/
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 1010;
+
+int n, m;
+int p[N];
+int d[N];
+
+int find(int x) {
+    if (p[x] != x) p[x] = find(p[x]);
+    return p[x];
+}
+
+int main() {
+#ifndef ONLINE_JUDGE
+    freopen("HDU1878.in", "r", stdin);
+#endif
+    while (~scanf("%d%d", &n, &m) && n) {
+        memset(d, 0, sizeof d);
+        for (int i = 1; i <= n; i++) p[i] = i;
+
+        while (m--) {
+            int a, b;
+            scanf("%d%d", &a, &b);
+            d[a]++, d[b]++;
+            p[find(a)] = find(b);
+        }
+
+        int cnt = 0;
+        for (int i = 1; i <= n; i++)
+            if (p[i] == i) cnt++;
+
+        int odd = 0;
+        for (int i = 1; i <= n; i++)
+            if (d[i] & 1) odd++;
+
+        if (cnt == 1 && odd == 0)
+            puts("1"); // 连通并且没有奇度点
+        else
+            puts("0");
+    }
+    return 0;
+}
+```