Codeforces Beta Round #14 (Div. 2) Two Paths （树形DP）

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 ai, bi (1 ≤ ai, bi ≤ 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
【题意】给你一棵树，要求找到两条链，使得乘积最大。
【分析】树形DP，对于每个节点维护两条最长链即可。

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 1e5+5;;
const int M = 17;
const int mod = 1e9+7;
const int mo=123;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int n,s;
int x[N],y[N];
vector<int>edg[N];
int dfs(int u,int fa){
    int mx1=0,mx2=0,ret=0;
    for(int v : edg[u]){
        if(v==fa)continue;
        ret=max(ret,dfs(v,u));
        if(s>mx1)mx2=mx1,mx1=s;
        else mx2=max(mx2,s);
    }
    ret=max(ret,mx1+mx2);
    s=mx1+1;
    return ret;
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<n;i++){
        scanf("%d%d",&x[i],&y[i]);
        edg[x[i]].pb(y[i]);
        edg[y[i]].pb(x[i]);
    }
    int ans=0;
    rep(i,1,n-1){
        int a=dfs(x[i],y[i]);
        int b=dfs(y[i],x[i]);
        ans=max(ans,a*b);
    }
    printf("%d\n",ans);
    return 0;
}