Two Paths--cf14D（树的直径）

题目链接：http://codeforces.com/problemset/problem/14/D

D. Two Paths

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

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.

Examples

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

直接暴力

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#include <vector>
#include <queue>

using namespace std;
#define memset(a,b) memset(a,b,sizeof(a))
#define N 250
int G[N][N],vis[N],dis[N];
int Index,Max,n;

void bfs(int u)
{
    Max=;
    memset(vis,);
    memset(dis,);
    queue<int>Q;
    Q.push(u);
    vis[u]=;
    while(!Q.empty())
    {
        int p;
        p=Q.front();
        Q.pop();
        for(int i=; i<=n; i++)
        {
            if(G[p][i]!=)
            {
                if(!vis[i])
                {
                    vis[i]=;
                    Q.push(i);
                    dis[i]=dis[p]+;
                    if(Max<dis[i])
                    {
                        Max=dis[i];
                        Index=i;
                    }
                }
            }
        }
    }
}

int main()
{
    int a[N],b[N];
    while(scanf("%d",&n)!=EOF)
    {
        memset(G,);
        memset(a,);
        memset(b,);
        for(int i=; i<n; i++)
        {
            scanf("%d %d",&a[i],&b[i]);
            G[a[i]][b[i]]=G[b[i]][a[i]]=;
        }
        int MM=;
        for(int i=; i<n; i++)
        {
            int x,y;
            G[a[i]][b[i]]=G[b[i]][a[i]]=;
            Index=;
            bfs(a[i]);
            bfs(Index);
            x=Max;

            Index=;
            bfs(b[i]);
            bfs(Index);
            y=Max;
            MM=max(MM,x*y);
            G[a[i]][b[i]]=G[b[i]][a[i]]=;
        }
        printf("%d\n",MM);
    }
    return ;
}