codeforces 14D 树的直径

题目大意：给你一棵树，要进行两次访问，两次访问中不能重复访问任一节点和边，问两次访问的最大长度乘积，每边 长度为1

思路：因为n只有200，所以可以枚举从哪里将树分成两棵树即删除一条边，求两棵树的直径，维护乘积最大值即可。树的直径在以前的博文讲过，两遍dfs即可

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <iomanip>

using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000")
#define maxn 205
#define MOD 1000000007
#define mem(a , b) memset(a , b , sizeof(a))
#define LL long long
#define ULL long long
const long long INF=0x3fffffff;
int a[maxn][maxn];
int vis[maxn] , n , maxx , ed;
struct node
{
    int u , v;
}edge[maxn];

void dfs(int u , int dis)
{
    vis[u] = 1;
    if(maxx <= dis)
    {
        maxx = dis;
        ed = u;
    }
    for(int i = 1 ; i <= n ; i ++ )
    {
        if(a[u][i] && !vis[i])
            dfs(i , dis + 1);
    }
}

int main()
{
    while(scanf("%d" , &n) != EOF)
    {
        int pos = 2 , u , v ;
        mem(a , 0);
        mem(vis , 0);
        for(int i = 2 ;i <= n ; i ++)
        {
            scanf("%d %d" , &u , &v);
            edge[pos].u = u;
            edge[pos++].v = v;
            a[u][v] = a[v][u] = 1;
        }
        int res = 0 , len1 , len2;
        for(int i = 2 ; i <= n ; i ++)
        {
            u = edge[i].u;
            v = edge[i].v;
            a[u][v] = a[v][u] = 0;
            mem(vis , 0);
            ed = -1 , maxx = 0;
            dfs(u , 0);
            mem(vis , 0);
            maxx = 0;
           // if(ed != -1)
            dfs(ed , 0);
            len1 = maxx;

            mem(vis , 0);
            ed = -1 , maxx = 0;
            dfs(v , 0);
            mem(vis , 0);
            maxx = 0;
          //  if(ed != -1)
            dfs(ed , 0);
            len2 = maxx;
            res = max(res , len1 * len2);
            a[u][v] = a[v][u] = 1;
        }
        printf("%d\n" , res);
    }
    return 0;
}

/*
15
13 14
10 14
5 10
10 6
9 10
10 7
15 6
8 7
2 6
1 10
3 1
3 11
4 3
14 12

*/