[POJ1655] Balancing Act

描述 Description

Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1…N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the forest T created by deleting that node from T.

For example, consider the tree:

Deleting node 4 yields two trees whose member nodes are {5} and {1,2,3,6,7}. The larger of these two trees has five nodes, thus the balance of node 4 is five. Deleting node 1 yields a forest of three trees of equal size: {2,6}, {3,7}, and {4,5}. Each of these trees has two nodes, so the balance of node 1 is two.

For each input tree, calculate the node that has the minimum balance. If multiple nodes have equal balance, output the one with the lowest number.

输入格式 InputFormat

The first line of input contains a single integer t (1 <= t <= 20), the number of test cases. The first line of each test case contains an integer N (1 <= N <= 20,000), the number of congruence. The next N-1 lines each contains two space-separated node numbers that are the endpoints of an edge in the tree. No edge will be listed twice, and all edges will be listed.

输出格式 OutputFormat

For each test case, print a line containing two integers, the number of the node with minimum balance and the balance of that node.

样例输入 SampleInput

1
7
2 6
1 2
1 4
4 5
3 7
3 1

样例输出 SampleOutput

1 2

POJ 1655

代码 Code

找树的重心和去掉重心之后子树最多的节点数。

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf = 0x7fffffff / 27.11;
const int maxn = 20005;
const int maxm = maxn * 2;
int i, j, t, n, m, l, r, k, z, y, x;
struct edge
{
    int to, nx;
} e[maxm];
int head[maxn], used[maxn], son[maxn];
int cnt, num, T, ans, sum;
inline void ins(int u, int v)
{
    e[++cnt] = (edge)
    {
        v, head[u]
    }; head[u] = cnt;
}
void dfs(int u, int fa, int tim)
{
    int i, t, v;
    son[u] = 1;
    used[u] = tim;
    for (t = 0, i = head[u]; i != 0; i = e[i].nx)
    {
        v = e[i].to;
        if (used[v] != tim && v != fa)
        {
            dfs(v, u, tim);
            son[u] += son[v];
            t = max(t, son[v]);
        }
    }
    t = max(t, n - son[u]);
    if ((t < sum) || (t == sum && ans > u))
    {
        ans = u;
        sum = t;
    }
}
inline void getcenter(int s)
{
    ans = 0; sum = inf;
    dfs(s, -1, ++num);
    printf("%d %d\n", ans, sum);
}
int main()
{
    num = 0;
    scanf("%d", &T);
    while (T--)
    {
        cnt = 0;
        memset(head, 0, sizeof(head));
        memset(son, 0, sizeof(son));
        scanf("%d", &n);
        for (i = 1; i < n; i++)
        {
            scanf("%d%d", &x, &y);
            ins(x, y); ins(y, x);
        }
        getcenter(1);
    }
    return 0;
}