题意:完全符合树的重心:即找到一个点,其所有的子树中最大的子树节点最少.
代码如下:
#include <cstdlib> #include <cstring> #include <cstdio> #include <iostream> #include <algorithm> #define MAXN 20000 using namespace std; //说白了这一题就是求一棵树的重心 int N, idx; // 说明有N个节点 //由于节点较多,且数为稀疏图,因此采用邻接表的形式来存储 struct Node { int x, next, cnt; // cnt用来说明该边所连接节点为子树的总节点数 bool vis; }e[(MAXN<<1)+5]; // 双向边所致 int head[MAXN+5]; void addedge(int a, int b) { ++idx; e[idx].vis = false, e[idx].cnt = 0; e[idx].x = b, e[idx].next = head[a]; head[a] = idx; } int dfs(int x) { // 得到某条边连接的子树的节点数(入口边不进行计算,否则全为N了) int sum = 0; for (int i = head[x]; i != -1; i = e[i].next) { if (!e[i].vis) { // 如果这条边没有被访问过 e[i].vis = e[i^1].vis = true; e[i].cnt += dfs(e[i].x); e[i^1].cnt = N-e[i].cnt; } sum += e[i].cnt; } return sum + 1; } void solve(int &x, int &y) { y = 0x7fffffff; for (int i = 1; i <= N; ++i) { int Max = 0x7fffffff+1; for (int k = head[i]; k != -1; k = e[k].next) { Max = max(Max, e[k].cnt); } if (Max < y) { y = Max; x = i; } } } int main() { int T; scanf("%d", &T); while (T--) { int a, b, x, y; memset(head, 0xff, sizeof (head)); idx = -1; scanf("%d", &N); for (int i = 1; i < N; ++i) { scanf("%d %d", &a, &b); addedge(a, b); addedge(b, a); } dfs(1); solve(x, y); printf("%d %d\n", x, y); } return 0; }