'commit'

TangDou/Topic/HuanGenDp/CF1406.cpp

@@ -0,0 +1,62 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 1e5 + 10;
+#define int long long
+#define endl "\n"
+
+vector<int> g[N]; // 邻接表，存图
+int sz[N];        // sz[i]:以i为根的子树中节点个数
+int son[N];       // son[i]:去掉节点i后，剩下的连通分量中最大子树节点个数
+int r1, r2, n;
+
+void dfs(int u, int fa) {
+    sz[u] = 1;  // u为根的子树中，最起码有一个节点u
+    son[u] = 0; // 把节点u去掉后，剩下的连通分量中最大子树节点个数现在还不知道，预求最大，先设最小
+
+    for (int i = 0; i < g[u].size(); i++) { // 枚举u的每一条出边
+        int v = g[u][i];
+        if (v == fa) continue;
+        dfs(v, u);
+        sz[u] += sz[v];
+        son[u] = max(son[u], sz[v]);
+    }
+    son[u] = max(son[u], n - sz[u]);
+    if ((son[u] << 1) <= n) r2 = r1, r1 = u;
+}
+
+signed main() {
+    int T;
+    cin >> T;
+    while (T--) {
+        cin >> n;
+        // 多组测试数据，清空
+        memset(sz, 0, sizeof sz);
+        memset(son, 0, sizeof son);
+        r1 = r2 = 0;
+
+        for (int i = 0; i <= n + 10; i++) g[i].clear();
+        for (int i = 1; i < n; i++) { // n-1条边
+            int x, y;
+            cin >> x >> y;
+            g[x].push_back(y);
+            g[y].push_back(x);
+        }
+
+        dfs(1, 0); // 以1号点为入口，它的父节点是0
+
+        if (!r2) {
+            int r3 = g[r1][0];
+            cout << r1 << " " << r3 << endl;
+            cout << r1 << " " << r3 << endl;
+        } else {
+            int r3 = r1;
+            for (int i = 0; i < g[r2].size(); i++) {
+                r3 = g[r2][i];
+                if (r3 != r1) break;
+            }
+            cout << r3 << " " << r2 << endl;
+            cout << r3 << " " << r1 << endl;
+        }
+    }
+    return 0;
+}

TangDou/Topic/HuanGenDp/CF1406C.cpp

@@ -1,71 +0,0 @@
-#include <bits/stdc++.h>
-#define N 100005
-using namespace std;
-int n, head[N], ver[2 * N], Next[2 * N], tot = 0;
-void add(int x, int y) {
-    ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
-}
-vector<int> c;
-int mx[N], sz[N];
-int mn = 0x3f3f3f3f;
-void dfs(int x, int pre) {
-    for (int i = head[x]; i; i = Next[i]) {
-        int y = ver[i];
-        if (y == pre) continue;
-        dfs(y, x);
-        sz[x] += (sz[y] + 1);
-        mx[x] = max(mx[x], sz[y] + 1);
-    }
-    mx[x] = max(mx[x], n - 1 - sz[x]);
-    mn = min(mn, mx[x]);
-}
-void solve() {
-    cin >> n;
-    tot = 0;
-    mn = 0x3f3f3f3f;
-    c.clear();
-    for (int i = 0; i <= n; i++) {
-        head[i] = 0;
-        mx[i] = 0;
-        sz[i] = 0;
-    }
-    for (int i = 1; i < n; i++) {
-        int x, y;
-        cin >> x >> y;
-        add(x, y);
-        add(y, x);
-    }
-    dfs(1, 0);
-    for (int i = 1; i <= n; i++) {
-        if (mx[i] == mn) {
-            c.push_back(i);
-        }
-    }
-    if (c.size() == 1) {
-        for (int i = head[c[0]]; i; i = Next[i]) {
-            int y = ver[i];
-            cout << c[0] << " " << y << endl;
-            cout << c[0] << " " << y << endl;
-            return;
-        }
-    } else {
-        int tmp;
-        for (int i = head[c[0]]; i; i = Next[i]) {
-            int y = ver[i];
-            if (y == c[1]) continue;
-            cout << c[0] << " " << y << endl;
-            tmp = y;
-            break;
-        }
-        cout << c[1] << " " << tmp << endl;
-        return;
-    }
-}
-int main() {
-    int T;
-    cin >> T;
-    while (T--) {
-        solve();
-    }
-    return 0;
-}