[POI2008]STA-Station

嘟嘟嘟

一道树形dp题。

令dp[u]表示以u为根时所有点的深度之和。考虑u到他的一个子节点v时答案的变化，v子树以外的点的深度都加１，v子树以内的点的深度都减１，所以dp[v] = dp[u] + (n - siz[v]) - siz[v]。于是dp式就搞出来了。

所以两边dfs，第一遍求siz和dp[1]，第二遍更新答案。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 1e6 + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) x = -x, putchar('-');
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

 int n, ans = ;
 struct Edge
 {
     int nxt, to;
 }e[maxn << ];
 int head[maxn], ecnt = ;
 void addEdge(int x, int y)
 {
     e[++ecnt] = (Edge){head[x], y};
     head[x] = ecnt;
 }

 int siz[maxn], dep[maxn];
 ll dp[maxn];
 void dfs1(int now, int f)
 {
     siz[now] = ;
     for(int i = head[now]; i; i = e[i].nxt)
     {
         if(e[i].to == f) continue;
         dep[e[i].to] = dep[now] + ;
         dp[] += dep[e[i].to];
         dfs1(e[i].to, now);
         siz[now] += siz[e[i].to];
     }

 }
 void dfs2(int now, int f)
 {
     for(int i = head[now]; i; i = e[i].nxt)
     {
         if(e[i].to == f) continue;
         dp[e[i].to] = dp[now] + n - (siz[e[i].to] << );
         if(dp[e[i].to] > dp[ans]) ans = e[i].to;
         dfs2(e[i].to, now);
     }
 }

 int main()
 {
     n = read();
     for(int i = ; i < n; ++i)
     {
         int x = read(), y = read();
         addEdge(x, y); addEdge(y, x);
     }
     dfs1(, ); dfs2(, );
     write(ans); enter;
     return ;
 }