codeforces --- 14D

C - Two Paths

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 14D

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

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAX = 205;
int dep, n, map[MAX][MAX];
int dfs(int u, int v){
    int max1, max2, path_max;
    max1 = max2 = path_max = 0;
    for(int i = 1;i <= n;i ++){
        if(map[i][u] && i != v){
            int z = dfs(i, u);
            if(path_max < z) path_max = z;
            if(max1 < dep){
                max2 = max1;
                max1 = dep;
            }
            else if(max2 < dep) max2 = dep;
        }
    }
    if(path_max < max1 + max2) path_max = max2 + max1;
    dep = max1 + 1;
    return path_max;
}

int main(int argc, char const *argv[]){
    int u, v;
    //freopen("in.c", "r", stdin);
    while(~scanf("%d", &n)){
        memset(map, 0, sizeof(map));
        for(int i = 0; i < n - 1;i ++){
            scanf("%d%d", &u, &v);
            map[u][v] = map[v][u] = 1;
        }
        int ans = -1;
        for(int i = 1;i <= n;i ++){
            for(int j = 1;j <= n;j ++){
                if(map[i][j]){
                    int a = dfs(i, j);
                    int b = dfs(j, i);
                    ans = max(ans, a*b);
                }
            }
        }
        printf("%d
", ans);
    }
    return 0;
}