HDU 3534 Tree (经典树形dp)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3534

题意：

给你一棵树，问你有多少对点的距离等于树的直径。

思路：

dp[i][0]表示在i的子树中 离i最远的距离，dp[i][1]是次远距离。   cnt[i][0]则是最远的点的数量，cnt[i][1]表示次远的数量。

up[i]表示以i向上 离i最远的距离。   up_cnt[i]表示向上最远的数量。

写的有点麻烦，调试了2小时。。。

 //#pragma comment(linker, "/STACK:102400000, 102400000")
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 typedef long long LL;
 typedef pair <int, int> P;
 const int N = 2e5 + ;
 int dp[N][];
 int cnt[N][];
 int up[N];
 int cnt_up[N];
 int son[N][];
 vector <P> G[N];

 void dfs1(int u, int p) {
     dp[u][] = dp[u][] = ;
     son[u][] = son[u][] = u;
     cnt[u][] = cnt[u][] = ;
     int tmp = ;
     for(int i = ; i < G[u].size(); ++i) {
         P temp = G[u][i];
         int v = temp.first;
         if(v == p)
             continue;
         dfs1(v, u);
         if(dp[v][] + temp.second > dp[u][]) {
             dp[u][] = dp[u][];
             son[u][] = son[u][];
             cnt[u][] = cnt[u][];
             dp[u][] = dp[v][] + temp.second;
             son[u][] = v;
             cnt[u][] = cnt[v][];
             tmp = cnt[v][];
         } else if(dp[v][] + temp.second == dp[u][]) {
             cnt[u][] += cnt[v][];
             cnt[u][] = cnt[u][] - tmp;
             dp[u][] = dp[u][];
             son[u][] = v;
         } else if(dp[v][] + temp.second > dp[u][]) {
             dp[u][] = dp[v][] + temp.second;
             cnt[u][] = cnt[v][];
             son[u][] = v;
         } else if(dp[v][] + temp.second == dp[u][]) {
             cnt[u][] += cnt[v][];
         }
     }
 }

 void dfs2(int u, int p) {
     for(int i = ; i < G[u].size(); ++i) {
         P temp = G[u][i];
         int v = temp.first;
         if(v == p)
             continue;
         if(dp[u][] == temp.second + dp[v][]) {
             //up[v] = max(up[u], dp[u][0]) + temp.second;
             if(dp[u][] == ) {
                 cnt_up[v] = cnt_up[u];
                 up[v] = up[u] + temp.second;
                 dfs2(v, u);
                 continue;
             }
             if(up[u] > dp[u][]) {
                 up[v] = up[u] + temp.second;
                 cnt_up[v] = cnt_up[u];
             } else if(dp[u][] > up[u]) {
                 up[v] = dp[u][] + temp.second;
                 if(dp[u][] == dp[u][])
                     cnt_up[v] = cnt[u][] - cnt[v][];
                 else
                     cnt_up[v] = cnt[u][];
             } else {
                 if(dp[u][] == dp[u][])
                     cnt_up[v] = cnt[u][] - cnt[v][];
                 else
                     cnt_up[v] = cnt[u][];
                 cnt_up[v] += cnt_up[u];
                 up[v] = dp[u][] + temp.second;
             }
         } else {
             //up[v] = max(up[u], dp[u][1]) + temp.second;
             if(up[u] > dp[u][]) {
                 up[v] = up[u] + temp.second;
                 cnt_up[v] = cnt_up[u];
             } else if(dp[u][] > up[u]) {
                 up[v] = dp[u][] + temp.second;
                 cnt_up[v] = cnt[u][];
             } else {
                 cnt_up[v] = cnt_up[u] + cnt[u][];
                 up[v] = dp[u][] + temp.second;
             }
         }
         dfs2(v, u);
     }
 }

 int main()
 {
     int n, u, v, c;
     while(~scanf("%d", &n)) {
         for(int i = ; i <= n; ++i) {
             G[i].clear();
         }
         for(int i = ; i < n; ++i) {
             scanf("%d %d %d", &u, &v, &c);
             G[u].push_back(make_pair(v, c));
             G[v].push_back(make_pair(u, c));
         }
         dfs1(, -);
         cnt_up[] = ;
         dfs2(, -);
         int Max = ;
         for(int i = ; i <= n; ++i) {
             //cout << dp[i][0] << "  -  " << up[i] << endl;
             Max = max(Max, max(dp[i][], up[i]));
         }
         LL ans = ;
         for(int i = ; i <= n; ++i) {
             if(Max == dp[i][]) {
                 ans += (LL)cnt[i][];
             }
             if(Max == up[i]) {
                 ans += (LL)cnt_up[i];
             }
         }
         printf("%d %lld\n", Max, ans/);
     }
     return ;
 }