Codeforces Round #14 D. Two Paths(求树上两条不相交的路径的乘积最大值)

题目链接：



http://codeforces.com/problemset/problem/14/D

思路：直接枚举每一天路径的两端，然后求以每一端为树根的树上最长路径，然后相乘就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define REP(i, a, b) for (int i = (a); i < (b); ++i)
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
using namespace std;

const int MAX_N = (200 + 20);
int N, ed, maxlen, ans, dp[MAX_N], pre[MAX_N];
vector<int > g[MAX_N];

void dfs(int u, int fa, int len, int limit)
{
	dp[u] = len;
	pre[u] = fa;
	REP(i, 0, (int)g[u].size()) {
		int v = g[u][i];
		if (v != fa && v != limit) dfs(v, u, len + 1, limit);
	}
}

int gao(int u, int v)
{
	memset(dp, 0, sizeof(dp));
	dfs(u, -1, 0, v);
	maxlen = 0; ed = u;
	FOR(i, 1, N) if (dp[i] > maxlen) maxlen = dp[i], ed = i;
	dfs(ed, -1, 0, v);
	maxlen = 0;
	FOR(i, 1, N) if (dp[i] > maxlen) maxlen = dp[i], ed = i;
	maxlen = 0;
	while (pre[ed] != -1) ed = pre[ed], ++maxlen;
	return maxlen;
}
int main()
{
	cin >> N;
	REP(i, 1, N) {
		int u, v; cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	ans = 0;
	FOR(i, 1, N) {
		REP(j, 0, (int)g[i].size()) {
			int l = gao(i, g[i][j]);
			int r = gao(g[i][j], i);
			ans = max(ans, l * r);
		}
	}
	cout << ans << endl;
	return 0;
}