Atcoder Grand Contest 010 C - Cleaning 树贪心（伪）

C - Cleaning

题目连接：

http://agc010.contest.atcoder.jp/tasks/agc010_c

Description

There is a tree with N vertices, numbered 1 through N. The i-th of the N−1 edges connects vertices ai and bi.

Currently, there are Ai stones placed on vertex i. Determine whether it is possible to remove all the stones from the vertices by repeatedly performing the following operation:

Select a pair of different leaves. Then, remove exactly one stone from every vertex on the path between those two vertices. Here, a leaf is a vertex of the tree whose degree is 1, and the selected leaves themselves are also considered as vertices on the path connecting them.
Note that the operation cannot be performed if there is a vertex with no stone on the path.

Input

The input is given from Standard Input in the following format:

N
A1 A2 … AN
a1 b1
:
aN−1 bN−1
2≦N≦105
1≦ai,bi≦N
0≦Ai≦109
The given graph is a tree.

Output

If it is possible to remove all the stones from the vertices, print YES. Otherwise, print NO.

Sample Input

5
1 2 1 1 2
2 4
5 2
3 2
1 3

Sample Output

YES

Hint

题意

给你一棵树，你每次可以选择两个叶子节点，使得这条路径上的所有点的点权减1，问你能否全部变成0.

题解：

考虑只有一层的时候，即一个点和一堆叶子，需要满足哪些条件，才能使得所有叶子节点的权值为0呢：

1.叶子权值和一定要大于等于父亲节点的权值，因为这样父亲节点才能满足下面的叶子节点的消耗。

2.叶子权值和的两倍要小于等于父亲节点的权值，因为叶子节点的权值每次是-2的，而父亲节点是-1.

3.叶子权值的最大值应该小于等于父亲节点。

根据这三个规则，一直递归的使得底层的点处理完之后，等价的看为叶子节点，然后不停跑就好了，有点树形dp的感觉……

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+6;
int n,a[maxn];
vector<int> E[maxn];
void dfs(int x,int p){
    if(E[x].size() == 1)return;
    long long sum = 0;
    long long mx = 0;
    for(int i=0;i<E[x].size();i++){
        int v = E[x][i];
        if(v==p)continue;
        dfs(v, x);
        sum += a[v];
        mx = max(1ll*a[v], mx);
    }
    if(a[x]>sum||sum>2*a[x]){
        cout<<"NO"<<endl;
        exit(0);
    }
    int k=sum-a[x];
    if(k>sum-mx){
        cout<<"NO"<<endl;
        exit(0);
    }
    a[x]-=k;
}
int main()
{
    scanf("%d", &n);
    for(int i=0;i<n;i++)
        scanf("%d", &a[i]);
    for(int i=1;i<n;i++){
        int x,y;
        scanf("%d%d", &x, &y);
        x--,y--;
        E[x].push_back(y);
        E[y].push_back(x);
    }
    if(n==2){
        if(a[0]==a[1])puts("YES");
        else puts("NO");
        return 0;
    }
    int v=0;
    while(E[v].size() == 1) v++;
    dfs(v, -1);
    if(a[v]==0)cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
}