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 = 200 + 10;

int n;

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

int ecnt;
int head[maxn];
Edge edges[maxn * 2];

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 + 1);
    }
}

int main()
{
    memset(head, -1, sizeof(head));

    scanf("%d", &n);
    for(int u, v, i = 1; i < n; i++)
    {
        scanf("%d%d", &u, &v);
        AddEdge(u, v); AddEdge(v, u);
    }

    int ans = 0;

    for(int i = 0; i < ecnt; i += 2)
    {
        int u = edges[i].u, v = edges[i].v;
        edges[i].del = true;
        edges[i^1].del = true;

        int t1, t2;

        id = u;
        len = 0;
        dfs(u, 0, 0);
        len = 0;
        dfs(id, 0, 0);
        t1 = len;

        id = v;
        len = 0;
        dfs(v, 0, 0);
        len = 0;
        dfs(id, 0, 0);
        t2 = len;

        ans = max(ans, t1 * t2);
        edges[i].del = false;
        edges[i^1].del = false;
    }

    printf("%d
", ans);

    return 0;
}