CodeForces 14D Two Paths

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 ton. 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), wheren is the amount of cities in the country. The followingn - 1 lines contain the information about the roads. Each line contains a pair of numbers of the cities, connected by the roada_i, b_i (1 ≤ a_i, b_i ≤ n).

Output

Output the maximum possible profit.

Sample test(s)

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

这题研究了好久，最后看了看别人的题解才过了。可能还是自己对树这一块布太熟悉吧，要多学习学习。

题目大意：

把一个图沿一条边割开，分成两个树，求这两个数中最长的路的乘积。

这里需要分别求两个树的直径。

因为题目数据不大，直接枚举边即可。

DFS从当前点开始搜索，把枚举的那条边先当作删掉，DFS当前点存在的最大深度和次大深度，两个相加不断更新最值，便可得到一边树的直径。

最后相乘即可。

#include <stdio.h>
#include <vector>
#include <algorithm>

using namespace std;
const int MAXN=205;
vector<int> graph[MAXN];

int currentSum=0;

int dfs(int startPoint,int endPoint)
{
        if(graph[startPoint].size()==0)
                return 0;
        int sum=0,firstMax=0,secondMax=0;
        for(unsigned int i=0;i<graph[startPoint].size();i++)
        {
                if(graph[startPoint][i]!=endPoint)
                {
                        int temp=dfs(graph[startPoint][i],startPoint);
                        sum=max(temp,sum);
                        if(currentSum>firstMax)
                        {
                                secondMax=firstMax;
                                firstMax=currentSum;
                        }
                        else if(currentSum>secondMax)
                                secondMax=currentSum;
                }
        }
        currentSum=firstMax+1;
        sum=max(sum,firstMax+secondMax);
        return sum;
}

int main()
{
        int n,a,b,ans=0;
        scanf("%d",&n);
        for(int i=0;i<n-1;i++)
        {
                scanf("%d%d",&a,&b);
                graph[a].push_back(b);
                graph[b].push_back(a);
        }
        for(int i=1;i<=n;i++)
                for(int j=0;j<graph[i].size();j++)
                {
                        int sumFromSourcei=dfs(i,graph[i][j]);
                        int sumFromSourcej=dfs(graph[i][j],i);
                        ans=max(ans,sumFromSourcei*sumFromSourcej);
                }
        printf("%d\n",ans);
        return 0;
}