# Codeforces beta round #69 (Div. 2 only) e question

Source: Internet
Author: User

Given a tree, each node has a weight. Given the starting point a, starting from, when each vertex is reached, a weight value of the vertex is obtained (the node is reduced by one) and the maximum number of values can be obtained...

Analysis: DFS increases the value of the root node by 1, and then performs the same operation on each subtree. Each time you select a subtree, you may get more priority options, at last, the parent node and the child node may have surplus and can be moved back and forth...
It is a tree-like DP .. Amount...

Code: well-written... Because a few int64s are missing, WA has a large film...

`# Include <iostream> # include <stdio. h >#include <cmath> # include <vector> # include <algorithm> using namespace STD; const int n = 100010; struct node {int P; _ int64 A, num; // maximum possibility of num subtree} TMP; int N, flag [N] ;__ int64 A [n], ans; vector <node> B [N]; int CMP (const node & A, const node & B) {return. num> B. num ;:__ int64 DFS (INT p) {vector <node >:: iterator it; int CNT = 0; For (IT = B [p]. begin (); it! = B [p]. end (); It ++) {If (flag [(* it ). p] = 0) {flag [(* it ). p] = 1; (* it ). num = DFS (* it ). p); flag [(* it ). p] = 0; (* it ). A = A [(* it ). p]; CNT ++ ;}}if (CNT = 0) {return 0 ;}sort (B [p]. begin (), B [p]. end (), CMP); _ int64 num = 0; A [p] --; For (IT = B [p]. begin (); A [p]> 0 & it! = B [p]. end () & (* it ). num> 0; it ++) {If (flag [(* it ). p] = 0) // (* it ). A is always greater than or equal to 1 .. {(* It ). A --; A [p] --; num + = (* it ). num + 2 ;}__ int64 num1 = 0; For (IT = B [p]. begin (); it! = B [p]. end (); It ++) {If (flag [(* it ). p] = 0 & (* it ). a> 0) {num1 + = (* it ). A ;}} num + = min (_ int64) A [p], num1) * 2; // A [p]-= min (_ int64) A [p], num1); // reversed... A [p] ++; return num;} int main () {int I, J, K, head; while (scanf ("% d", & N )! = EOF) {for (I = 1; I <= N; I ++) {scanf ("% i64d", & A [I]); flag [I] = 0 ;}for (I = 1; I <= n-1; I ++) {scanf ("% d", & J, & K ); TMP. P = K; TMP. A = A [k]; TMP. num = 0; B [J]. push_back (TMP); TMP. P = J; TMP. A = A [J]; TMP. num = 0; B [K]. push_back (TMP);} scanf ("% d", & head); A [head] ++; flag [head] = 1; ans = DFS (head ); printf ("% i64d \ n", ANS); for (I = 1; I <= N; I ++) {While (! B [I]. Empty () {B [I]. pop_back () ;}} return 0 ;}`
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