CF-13D - Two Paths（搜索）

最新推荐文章于 2018-12-12 15:33:00 发布

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本文深入探讨了深度学习和人工智能算法在不同领域的应用，包括自然语言处理、图像处理、音视频处理等，展示了这些技术如何解决实际问题，并提供了解决方案的详细步骤。

D - Two Paths

Crawling in process...Crawling failedTime 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

Output

Input

Output

Input

Output

思路：因为题目给的是一棵树求两条无公共点的最长路径的积。所以可以随便在树里选条边删掉，再剩下的两颗子树中找最长的相乘即可。

如图选删去AB边，那就从A开始搜两条最长分支的路径之和就是A所在部分的最长路径。同样搜B。

所以A所在的只有一条最长分支为3，B有两条，一个是3一个是2所以B所在最长为4

#include<iostream>
#include<cstring>
using namespace std;
const int mm=233;
int map[mm][mm];
int n,ans;
int dep;
int dfs(int u,int p)
{
  int max1=0,max2=0;
  int path_max=0;
  for(int i=1;i<=n;i++)
    if(map[u][i]&&i!=p)
  {
    int z=dfs(i,u);
    if(path_max<z)path_max=z;
    if(max1<dep)///更新当前点的两条最长路径
    {
      max2=max1;max1=dep;
    }
    else if(max2<dep)max2=dep;

  }
  if(path_max<max1+max2)path_max=max1+max2;///最长路径等于最长两条分路径和
  dep=max1+1;///更新上层最大深度为当前层最大深度
  return path_max;
}

int main()
{
  int a,b;
  while(cin>>n)
  { memset(map,0,sizeof(map));
    for(int i=0;i<n-1;i++)
    {
      cin>>a>>b;map[a][b]=map[b][a]=1;
    }
    ans=0;
    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);
      if(ans<a*b)
        ans=a*b;
    }
    cout<<ans<<"\n";
  }
}