CodeForces 14D 树的直径 Two Paths

给出一棵树，找出两条不相交即没有公共点的路径，使得两个路径的长度的乘积最大。

思路：枚举树中的边，将该边去掉，分成两棵树，分别求出这两棵树的直径，乘起来维护一个最大值即可。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;

 const int maxn =  + ;

 int n;

 struct Edge
 {
     int u, v, nxt;
     bool del;
 };

 int ecnt;
 int head[maxn];
 Edge edges[maxn * ];

 void AddEdge(int u, int v)
 {
     edges[ecnt].u = u;
     edges[ecnt].v = v;
     edges[ecnt].nxt = head[u];
     edges[ecnt].del = false;
     head[u] = ecnt++;
 }

 int len, id;

 void dfs(int u, int fa, int dep)
 {
     if(dep > len) { len = dep; id = u; }
     for(int i = head[u]; ~i; i = edges[i].nxt)
     {
         if(edges[i].del) continue;
         int v = edges[i].v;
         if(v == fa) continue;
         dfs(v, u, dep + );
     }
 }

 int main()
 {
     memset(head, -, sizeof(head));

     scanf("%d", &n);
     for(int u, v, i = ; i < n; i++)
     {
         scanf("%d%d", &u, &v);
         AddEdge(u, v); AddEdge(v, u);
     }

     int ans = ;

     for(int i = ; i < ecnt; i += )
     {
         int u = edges[i].u, v = edges[i].v;
         edges[i].del = true;
         edges[i^].del = true;

         int t1, t2;

         id = u;
         len = ;
         dfs(u, , );
         len = ;
         dfs(id, , );
         t1 = len;

         id = v;
         len = ;
         dfs(v, , );
         len = ;
         dfs(id, , );
         t2 = len;

         ans = max(ans, t1 * t2);
         edges[i].del = false;
         edges[i^].del = false;
     }

     printf("%d\n", ans);

     return ;
 }

代码君