Codeforces Beta Round #69 (Div. 1 Only) C. Beavermuncher-0xFF 树上贪心

题目链接：

http://codeforces.com/problemset/problem/77/C

C. Beavermuncher-0xFF

time limit per test：3 seconds
memory limit per test：256 megabytes
#### 问题描述
> "Eat a beaver, save a tree!" — That will be the motto of ecologists' urgent meeting in Beaverley Hills.
>
> And the whole point is that the population of beavers on the Earth has reached incredible sizes! Each day their number increases in several times and they don't even realize how much their unhealthy obsession with trees harms the nature and the humankind. The amount of oxygen in the atmosphere has dropped to 17 per cent and, as the best minds of the world think, that is not the end.
>
> In the middle of the 50-s of the previous century a group of soviet scientists succeed in foreseeing the situation with beavers and worked out a secret technology to clean territory. The technology bears a mysterious title "Beavermuncher-0xFF". Now the fate of the planet lies on the fragile shoulders of a small group of people who has dedicated their lives to science.
>
> The prototype is ready, you now need to urgently carry out its experiments in practice.
>
> You are given a tree, completely occupied by beavers. A tree is a connected undirected graph without cycles. The tree consists of n vertices, the i-th vertex contains ki beavers.
>
> "Beavermuncher-0xFF" works by the following principle: being at some vertex u, it can go to the vertex v, if they are connected by an edge, and eat exactly one beaver located at the vertex v. It is impossible to move to the vertex v if there are no beavers left in v. "Beavermuncher-0xFF" cannot just stand at some vertex and eat beavers in it. "Beavermuncher-0xFF" must move without stops.
>
> Why does the "Beavermuncher-0xFF" works like this? Because the developers have not provided place for the battery in it and eating beavers is necessary for converting their mass into pure energy.
>
> It is guaranteed that the beavers will be shocked by what is happening, which is why they will not be able to move from a vertex of the tree to another one. As for the "Beavermuncher-0xFF", it can move along each edge in both directions while conditions described above are fulfilled.
>
> The root of the tree is located at the vertex s. This means that the "Beavermuncher-0xFF" begins its mission at the vertex s and it must return there at the end of experiment, because no one is going to take it down from a high place.
>
> Determine the maximum number of beavers "Beavermuncher-0xFF" can eat and return to the starting vertex.
#### 输入
> The first line contains integer n — the number of vertices in the tree (1 ≤ n ≤ 105). The second line contains n integers ki (1 ≤ ki ≤ 105) — amounts of beavers on corresponding vertices. Following n - 1 lines describe the tree. Each line contains two integers separated by space. These integers represent two vertices connected by an edge. Vertices are numbered from 1 to n. The last line contains integer s — the number of the starting vertex (1 ≤ s ≤ n).
#### 输出
> Print the maximum number of beavers munched by the "Beavermuncher-0xFF".
>
> Please, do not use %lld specificator to write 64-bit integers in C++. It is preferred to use cout (also you may use %I64d).
#### 样例
> **sample input**
> 5
> 1 3 1 3 2
> 2 5
> 3 4
> 4 5
> 1 5
> 4
>
> **sample output**
> 6

题意

给你一颗树，树上每个点有ai个小球，现在你有一个机器人位于根节点，这个机器人每走到一个节点，就会吃掉这个节点上的一个小球（初始位置一开始不会吃），如果一个位置没有小球，这个机器人是过不去的，而且这个机器人不会停下来，现在问你从根节点开始吃，最后要求能回到根节点，能吃到的最多小球。

题解

贪心：

策略1：从u进入一个子节点v的时候，优先考虑吃掉v的子树，再考虑u，v之间来回吃。

策略2：对于u的所有子节点vi，定义sumv[vi]为吃掉最多的以vi为根的子树的球（并且回到vi）。那么我们对于u，优先考虑先吃sumv[vi]比较大的子树。

代码

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn=1e5+10;
typedef __int64 LL;

//cntv[v]：表示v节点的小球个数
//sumv[v]：表示吃完以v为根的子树并回到v能吃到的最多小球。
//remv[v]：表示吃完以v为根的子树并回到v能吃到的最多小球之后，v这个节点还剩下的小球个数。
LL cntv[maxn],sumv[maxn],remv[maxn];
int n,rt;

void init(){
	memset(remv,0,sizeof(remv));
	memset(sumv,0,sizeof(sumv));
}

vector<int> G[maxn];
void dfs(int u,int fa){
	if(G[u].size()==1&&fa!=-1){
		remv[u]=cntv[u];
		sumv[u]=0;
	}else{
		priority_queue<LL> pq;
		for(int i=0;i<G[u].size();i++){
			int v=G[u][i];
			if(v==fa) continue;
			dfs(v,u);
			pq.push(sumv[v]);
		}
		if(fa!=-1){
			sumv[u]++; cntv[u]--;
		}
		while(cntv[u]&&!pq.empty()&&pq.top()){
			LL t=pq.top(); pq.pop();
			sumv[u]+=t;
			cntv[u]--;
			sumv[u]++;
		}
		for(int i=0;i<G[u].size()&&cntv[u];i++){
			int v=G[u][i];
			if(v==fa) continue;
			if(remv[v]==0) continue;
			if(cntv[u]>=remv[v]){
				sumv[u]+=2*remv[v];
				cntv[u]-=remv[v];
			}else{
				sumv[u]+=2*cntv[u];
				cntv[u]=0;
			}
		}
		remv[u]=cntv[u];
	}
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&cntv[i]);
	}
	for(int i=1;i<n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	scanf("%d",&rt);
	dfs(rt,-1);
	printf("%I64d\n",sumv[rt]);
	return 0;
}

/*
5
1 1 1 1 1
2 5
3 4
4 5
1 5
4

6
3 3 1 2 3 5
2 6
1 6
4 5
5 1
3 4
5
*/