Codeforces--14D--Two Paths（树的直径）



Two Paths

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit
Status

Description

As you know, Bob's brother lives in Flatland. In Flatland there are
n cities, connected by n - 1 two-way roads. The cities are numbered from 1 to
n. You can get from one city to another moving along the roads.

The «Two Paths» company, where Bob's brother works, has won a tender to repair two paths in Flatland. A path is a sequence of different cities, connected sequentially by roads. The company is allowed to choose by itself the paths to repair. The only condition
they have to meet is that the two paths shouldn't cross (i.e. shouldn't have common cities).

It is known that the profit, the «Two Paths» company will get, equals the product of the lengths of the two paths. Let's consider the length of each road equals 1, and the length of a path equals the amount of roads in it. Find the maximum possible profit
for the company.

Input

The first line contains an integer n (2 ≤ n ≤ 200), where
n is the amount of cities in the country. The following
n - 1 lines contain the information about the roads. Each line contains a pair of numbers of the cities, connected by the road
a_i, b_i (1 ≤ a_i, b_i ≤ n).

Output

Output the maximum possible profit.

Sample Input

Input

4
1 2
2 3
3 4

Output

1

Input

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

Output

0

Input

6
1 2
2 3
2 4
5 4
6 4

Output

4

Sample Output

Hint

Source

Codeforces Beta Round #14 (Div. 2)

n个点，n-1条无向边求这个图中最长的两条不相交的链长度，因为这里只有n-1条边，所以这一定不会存在环，所以我们可以枚举每一条边，用这条边把图分成两部分，求各自部分的树的直径，然后乘积，dfs的时候记录最长和次长的树链长度

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<int>G[1010];
int ans,n;
int dfs(int v,int x)
{
	int sum=0;
	int max1=0,max2=0;
	for(int i=0;i<G[v].size();i++)
	{
		if(G[v][i]!=x)
		{
			sum=max(sum,dfs(G[v][i],v));
			if(ans>max1)
			{
				max2=max1;
				max1=ans;
			}
			else
			max2=ans>max2?ans:max2;
		}
	}
	sum=max(sum,max1+max2);
	ans=max1+1;
	return sum;
}
int main()
{
	while(cin>>n)
	{
		for(int i=1;i<=n;i++)
		G[i].clear();
		int x,y;
		for(int i=0;i<n-1;i++)
		{
			ans=0;
			cin>>x>>y;
			G[x].push_back(y);
			G[y].push_back(x);
		}
		int maxx=0;
		for(int i=1;i<=n;i++)
		{
			for(int j=0;j<G[i].size();j++)
			{
				ans=0;
				x=dfs(G[i][j],i);
				y=dfs(i,G[i][j]);
				maxx=max(maxx,x*y);
			}
		}
		cout<<maxx<<endl;
	}
	return 0;
}