题目链接：点击打开链接

题意：给定一棵树

找2条点不重复的路径，使得两路径的长度乘积最大

思路：

1、为了保证点不重复，在图中删去一条边，枚举这条删边

2、这样得到了2个树，在各自的树中找最长链，即树的直径，然后相乘即可

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<set>
#include<vector>
#include<map>
#include<math.h>
#include<queue>
#include<string>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define N 220
struct Edge {
	int from, to, nex;
	bool hehe;
}edge[N<<1];
int head[N], edgenum;
void add(int u, int v){
	Edge E={u,v,head[u],true};
	edge[edgenum] = E;
	head[u] = edgenum ++;
}
int n;
int dis[N];
void init(){ memset(head, -1, sizeof head);	edgenum = 0; }
int bfs(int x){
	memset(dis, 0, sizeof dis);
	dis[x] = 1;
	queue<int>q; q.push(x);
	int hehe = x;
	while(!q.empty()) {
		int u = q.front(); q.pop();
		for(int i = head[u]; ~i; i = edge[i].nex) {
			if(!edge[i].hehe)continue;
			int v = edge[i].to;
			if(dis[v])continue;
			dis[v] = dis[u]+1, q.push(v);
			if(dis[hehe]<dis[v])hehe = v;
		}
	}
	return hehe;
}
int main(){
	int i, u, v;
	while(~scanf("%d",&n)){
		init();
		for(i=1;i<n;i++){
			scanf("%d %d",&u,&v);
			add(u,v); add(v,u);
		}
		int ans = 0;
		for(i = 0; i < edgenum; i+=2)
		{
			edge[i].hehe = edge[i^1].hehe = false;
			u = edge[i].from; v = edge[i].to;
			int L1 = bfs(u); int R1 = bfs(L1);
			int mul1 = dis[R1] -1;
			int L2 = bfs(v); int R2 = bfs(L2);
			int mul2 = dis[R2] -1;
			ans = max(ans, mul1 * mul2);
			edge[i].hehe = edge[i^1].hehe = true;
		}
		printf("%d\n",ans);
	}
	return 0;
}