SGU134 Centroid

题目大意

对于树上某个节点，若删除该点后得到的最大连通分量最小，这样的点称为树的重心
找出所有的重心以及对应的最小值

算法思路

DFS处理每棵子树的大小，
每个节点的值等于，下面最大的子树大小，以及整棵树删除该节点对应子树的大小中，较大的值

时间复杂度: O(N)

代码

/** * Copyright © 2015 Authors. All rights reserved. *  * FileName: 134.cpp * Author: Beiyu Li <[e-mail]> * Date: 2015-06-14 */#include <bits/stdc++.h>using namespace std;#define rep(i,n) for (int i = 0; i < (n); ++i)#define For(i,s,t) for (int i = (s); i <= (t); ++i)#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)typedef long long LL;typedef pair<int, int> Pii;const int inf = 0x3f3f3f3f;const LL infLL = 0x3f3f3f3f3f3f3f3fLL;const int maxn = 16000 + 5;int n, psz;struct Edge {        int v;        Edge *next;} epool[maxn*2], *e[maxn];int tot[maxn], val[maxn];void add_edge(int u, int v){        Edge *i = epool + psz++;        i->v = v; i->next = e[u]; e[u] = i;}void dfs(int u, int f){        tot[u] = 1;        for (Edge *i = e[u]; i; i = i->next) {                int v = i->v;                if (v == f) continue;                dfs(v, u);                tot[u] += tot[v];                val[u] = max(val[u], tot[v]);        }        val[u] = max(val[u], n - tot[u]);}int main(){        scanf("%d", &n);        rep(i,n-1) {                int u, v;                scanf("%d%d", &u, &v);                --u, --v;                add_edge(u, v);                add_edge(v, u);        }        dfs(0, -1);        int ret = *min_element(val, val + n), cnt = 0;        rep(i,n) if (val[i] == ret) ++cnt;        printf("%d %d\n", ret, cnt);        rep(i,n) if (val[i] == ret) printf("%d%c\n", i + 1, --cnt? ' ': '\n');        return 0;}

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