[POI2014]MRO-Ant colony

嘟嘟嘟

题面很迷，看这个吧。

首先暴力很简单，从每一个叶子节点开始爬，直到那条特殊的边。

正解稍微想想就能搞出来：(x, y)这条特殊的边把整棵树分成了两部分，然后我们分别从x, y开始在他的那部分子树dfs，求出到达节点v时满足条件的一个区间。因为从v到u是向下取整，那么反过来合法的区间就是[k * (du[u] - 1), k * (du[u] - 1) + (du[u] - 1) - 1]。

递归到叶子节点后在原序列中二分找在[Min, Max]中的数的个数，乘上k，累加到答案中。

时间复杂度O(nlogn)。

 #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 + ;
 const int MAX_NUM = 2e9;
 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, m, k;
 ll a[maxn];
 int s1, s2, du[maxn];
 ll 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;
 }

 void dfs(int now, int f, ll Min, ll Max)
 {
   if(du[now] == )
     {
       int l = lower_bound(a + , a + m + , Min) - a;
       int r = upper_bound(a + , a + m + , Max) - a - ;
       ans += (r - l + ) * k;
       return;
     }
   int x = du[now] - ;
   for(int i = head[now]; i != -; i = e[i].nxt)
     {
       if(e[i].to == f) continue;
       dfs(e[i].to, now, Min * x, Max * x + x - );
     }
 }

 int main()
 {
   Mem(head, -);
   n = read(); m = read(); k = read();
   for(int i = ; i <= m; ++i) a[i] = read();
   sort(a + , a + m + );
   for(int i = ; i < n; ++i)
     {
       int x = read(), y = read();
       du[x]++; du[y]++;
       if(i == ) s1 = x, s2 = y;
       addEdge(x ,y); addEdge(y, x);
     }
   dfs(s1, s2, k, k); dfs(s2, s1, k, k);
   write(ans), enter;
   return ;
 }