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为了解决这个问题，我们需要找到树形图中两个点，使得这两个点的病人数乘积最大。我们可以通过以下步骤来解决这个问题：
#include#include #include using namespace std;void dfs(int u, int parent, const vector >& adj, const vector & a, int& max_dist, int& far_node) { max_dist = 0; far_node = u; for (int v : adj[u]) { if (v == parent) continue; int current_dist = 1 + dfs(v, u, adj, a, max_dist, far_node); if (current_dist > max_dist) { max_dist = current_dist; far_node = v; } } } int main() { int n; vector a(n + 1); // 读取输入 // 假设输入的第一行是n，接下来的n行是a[1]到a[n] // 读取n-1条边，构建邻接表 vector > adj(n + 1); for (int i = 1; i <= n; ++i) { for (int j = 1; j <= n; ++j) { if (i == j) continue; adj[i].push_back(j); adj[j].push_back(i); } } // 第一次DFS找到最远的点u int u = 1; int max_dist = 0; for (int v = 1; v <= n; ++v) { if (v == u) continue; int current_dist; int far_node; dfs(v, -1, adj, a, current_dist, far_node); if (current_dist > max_dist) { max_dist = current_dist; u = far_node; } } // 第二次DFS从u出发找到最远的点v int v = 1; max_dist = 0; for (int w = 1; w <= n; ++w) { if (w == u) continue; int current_dist; int far_node; dfs(w, -1, adj, a, current_dist, far_node); if (current_dist > max_dist) { max_dist = current_dist; v = far_node; } } // 计算u和v的ax乘积，并检查是否存在更大的乘积 int max_product = a[u] * a[v]; for (int i = 1; i <= n; ++i) { for (int j = 1; j <= n; ++j) { if (i == j) continue; if (a[i] * a[j] > max_product) { max_product = a[i] * a[j]; } } } cout << max_product << endl; return 0; }
通过这种方法，我们可以高效地找到树中的最大乘积点对。
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