51Nod 1405 树的距离之和 (树dp)

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1405

中文题面不解释了，两次dfs，第一次自下向上，第二次自上向下。

ans[i]表示i节点的答案，cnt[i]表示i节点为root的子树的节点个数，d[i]表示i节点为root的子树的答案。

 //#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 = 1e5 + ;
 LL cnt[N], d[N], ans[N], n;
 vector <int> edge[N];

 void dfs1(int u, int p) {
     cnt[u] = ;
     d[u] = ;
     for(int i = ; i < edge[u].size(); ++i) {
         int v = edge[u][i];
         if(v == p)
             continue;
         dfs1(v, u);
         cnt[u] += cnt[v];
         d[u] += cnt[v] + d[v];
     }
 }

 void dfs2(int u, int p) {
     if(p != -) {
         ans[u] = (ans[p] - d[u] - cnt[u]) + (n - cnt[u]) + d[u];
     } else {
         ans[u] = d[u];
     }
     for(int i = ; i < edge[u].size(); ++i) {
         int v = edge[u][i];
         if(v == p)
             continue;
         dfs2(v, u);
     }
 }

 int main()
 {
     int u, v;
     while(~scanf("%lld", &n)) {
         for(int i = ; i <= n; ++i) {
             edge[i].clear();
         }
         for(int i = ; i < n; ++i) {
             scanf("%d %d", &u, &v);
             edge[u].push_back(v);
             edge[v].push_back(u);
         }
         dfs1(, -);
         dfs2(, -);
         for(int i = ; i <= n; ++i) {
             printf("%lld\n", ans[i]);
         }
     }
     return ;
 }