XMU1456 water body

1456. squirrel harvest Time Limit: 2000 MS Memory Limit: 65536 KTotal Submissions: 64 (23 users) Accepted: 21 (20 users) [My Solution] Description Squirrel lives in a Tree Containing n + 1 nodes. It lives in node 0. Of course, to survive, it will go out to collect results every day. To simplify the problem, we assume that pine nuts only appear on n other nodes except node 0. Squirrel starts from node 0 every day and returns a pine nut to a node, on the way back from the original road, take each passing node to pick up one pine cones (you can choose not to pick them ). So, Squirrel wants to know how many days later the fruit on the tree will be fully lit by it? The first line of Input has a positive integer n (n <= 50,000), and the next line has n integers (1 <= I <= n, 1 <= ai <= 20,000). the I-th number indicates that the fruit number on the node numbered I is ai, followed by n rows, each row has two integers x, y (0 <= x, y <= n) indicates that two nodes are connected by an edge. The input ensures that any two points in the given graph have only one path. Output outputs an integer that represents the number of days required. Sample Input3 3 2 2 0 1 1 2 3 3 Sample Output4Sourcedoraemon @ xmu [Submit] [Statistic] [Status] [Discuss] http://acm.xmu.edu.cn/JudgeOnline/problem.php? Id = 1456 question: A tree with n nodes starts from node 0 to each node to fetch fruit. Each time a tree with n nodes picks up a fruit and returns, the fruit of a passing node can also be picked. how many times can all the fruit on the tree be taken: the minimum number of times required for each node of a branch depends on the sum of the current node and all its subnodes. [cpp] # include <stdio. h ># include <vector> using namespace std; vector <int> a [50000 + 100]; int val [50000 + 100], vis [50000 + 100]; int DFS (int k) {if (vis [k]) return 0; vis [k] = 1; if (a [k]. size () = 0) return val [k]; int sum = 0; for (int I = 0; I <a [k]. size (); I ++) sum + = DFS (a [k] [I]); Return val [k]> sum? Val [k]: sum;} int main () {int n, I, x, y, ans; while (scanf ("% d", & n )! = EOF) {for (I = 0; I <50000 + 10; I ++) {a [I]. clear (); val [I] = 0; vis [I] = 0;} for (I = 1; I <= n; I ++) scanf ("% d", & val [I]); for (I = 1; I <= n; I ++) {scanf ("% d ", & x, & y); a [x]. push_back (y); a [y]. push_back (x);} ans = DFS (0); printf ("% d \ n", ans);} return 0 ;}