hihocoder 1224 赛车 （我只能说LCA，T了）

先说一下我用LCA的思路：输入过程中把叶子结点找出来。然后dfs找出各个点到根节点1的距离，并且记录下最长路的长度len和非根节点端点tail，而且要求出来两个端点都是叶子结点的LCA。

然后答案就是len+d[i]-d[lca(i,tail)]（i为其它叶子结点）中的最大值了。

下面是T了的代码o(╯□╰)o

#pragma warning(disable:4996)#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <map>using namespace std;const int N = 100005;int first[N], nxt[N], to[N], e;int d[N];//距离1的最距离bool vis[N];int father[N];int n;map<pair<int, int>, int>lca;vector<int>ans;int find_set(int x){if (x == father[x])return father[x];return father[x] = find_set(father[x]);}void dfs(int u, int dis){d[u] = dis;father[u] = u;for (int i = first[u]; i != -1; i = nxt[i]){int v = to[i];if (father[v] == -1){dfs(v, dis + 1);father[find_set(v)] = u;}}if (vis[u])return;for (int j = 0; j < (int)ans.size(); j++){int i = ans[j];if (father[i] != -1){lca[pair<int, int>(i, u)] = lca[pair<int, int>(u, i)] = find_set(i);}}}int main(){scanf("%d", &n);memset(vis, false, sizeof vis);memset(first, -1, sizeof first);e = 0;for (int i = 1; i < n; i++){int u, v; scanf("%d %d", &u, &v);nxt[e] = first[u];to[e] = v;first[u] = e++;vis[u] = true;}for (int i = 1; i <= n; i++){if (vis[i])continue;ans.push_back(i);}memset(father, -1, sizeof father);memset(d, -1, sizeof d);dfs(1, 0);int len = 0, tail = 1;for (int i = 1; i <= n; i++){if (len < d[i]){tail = i;len = d[i];}}int anss = len;for (int j = 0; j < (int)ans.size(); j++){int i = ans[j];if (vis[i] || i == tail)continue;int u = lca[pair<int, int>(i, tail)];int tmp = len + d[i] - d[u];anss = max(tmp, anss);}cout << anss << endl;return 0;}

然后说一下题解的思路吧，就是求出来dfs的时候顺便求出来每个点向下能到达的最大深度down数组。然后答案就是len+down[i]+1中的最大值了，注意下此处的 i 不能在最长路中要不会成环。。

下面是代码：

#pragma warning(disable:4996)#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 100005;int first[N], nxt[N], to[N], e;//邻接表int d[N];//距离节点1的最距离int down[N];//向下的最大距离int n;int fa[N];//记录父亲结点bool vis[N];//记录该节点是否在最长路上int tail = 0, len = 0;//记录最长路长度和端点void dfs(int u, int dis){down[u] = 0;d[u] = dis;if (dis > len){len = dis;tail = u;}for (int i = first[u]; i != -1; i = nxt[i]){int v = to[i];if (d[v] == -1){fa[v] = u;dfs(v, dis + 1);}down[u] = max(down[u], down[v] + 1);}}int main(){scanf("%d", &n);memset(first, -1, sizeof first);e = 0;for (int i = 1; i < n; i++){int u, v; scanf("%d %d", &u, &v);nxt[e] = first[u];to[e] = v;first[u] = e++;}memset(d, -1, sizeof d);memset(fa, -1, sizeof fa);dfs(1, 0);//沿着父亲节点标记处最长路上的点int last = tail;while (last != -1){vis[last] = true;last = fa[last];}int ans = len;for (int i = 1; i <= n; i++){if (vis[i])continue;ans = max(ans, len + down[i] + 1);}cout << ans << endl;return 0;}