题目连接:http://codeforces.com/contest/526/problem/B
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square . Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and . All numbers ai are positive integers, not exceeding 100.
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
2 1 2 3 4 5 6
5
Picture for the sample test. Green color denotes the additional street lights.
题目大意:
给定一个完全二叉树,各边有权值,要求使每颗子树的左右子树边权相同,问最少需要添加多少边权。
解题思路:
利用二叉树的性质,自低向上递推即可,每一层求出各个左右兄弟节点之差(当然求绝对值)求和,每个父亲节点在计算之前加上左右子节点中的最大值,以达到任意子树都相同的目的,推到第一层即可。
代码如下:
#includeusing namespace std;int tree[4100]= { 0};int main(){ int n; scanf("%d",&n); int s=1; for(int i=0; i<=n; i++) s*=2; for(int i=2; i<=s-1; i++) scanf("%d",&tree[i]); int ans=0; s/=2; while(s>1) { for(int i=s; i<=s*2-1; i++) tree[i]+=max(tree[i*2],tree[i*2+1]); for(int i=s; i<=s*2-2; i+=2) ans+=abs(tree[i]-tree[i+1]); s/=2; } cout< <