Codeforces Round #168 (Div. 2)D. Zero Tree（DP，中等难度）

D. Zero Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A tree is a graph with n vertices and exactly n - 1 edges; this graph should meet the following condition: there exists exactly one shortest (by number of edges) path between any pair of its vertices.

A subtree of a tree T is a tree with both vertices and edges as subsets of vertices and edges of T.

You're given a tree with n vertices. Consider its vertices numbered with integers from 1 to n. Additionally an integer is written on every vertex of this tree. Initially the integer written on the i-th vertex is equal to v_i. In one move you can apply the following operation:

1. Select the subtree of the given tree that includes the vertex with number 1.
2. Increase (or decrease) by one all the integers which are written on the vertices of that subtree.

Calculate the minimum number of moves that is required to make all the integers written on the vertices of the given tree equal to zero.

Input

The first line of the input contains n (1 ≤ n ≤ 10⁵). Each of the next n - 1 lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i) indicating there's an edge between vertices a_i and b_i. It's guaranteed that the input graph is a tree.

The last line of the input contains a list of n space-separated integers v₁, v₂, ..., v_n (|v_i| ≤ 10⁹).

Output

Print the minimum number of operations needed to solve the task.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Sample test(s)

Input

3
1 2
1 3
1 -1 1

Output

3

一开始题目理解错误：“includes the vertex with number 1.”是指结点1而非其上integer为1的结点。
思路是先处理儿子再处理父亲。这是因为：设有结点x，x的val变为0所需操作次数x.count = in + de + r,这里in是x因为儿子结点的+1操作
而随之产生的+1操作的次数（对儿子的操作会传递到父亲），而de就是-1操作，显然，in是x的所有儿子中+1次数的最大值，de是所有儿子中-1次
数的最大值。而r是什么？在处理完儿子后确保儿子的val为0，但不保证x的val为0，r就是余下的对x操作的次数：r=x.val+in-de（这些操作是
发生在x和他的父亲之间的）。由以上于可见，对x.count的统计能转化为对其儿子结点的相关信息的统计，故而此前说“先处理儿子后处理父亲”。
结点1是最后处理的。
这里还涉及如何保存一个结点的所有儿子的信息的问题，我开始用邻接表，超时，后来参考别人的代码才想起来可以用vector。

AC Code：

#include <iostream>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define cti const int
#define dg(i) cout << "*" << i << endl;

cti MAXN = 100005;
int n;
bool vis[MAXN];
LL v[MAXN];
struct Count
{
    LL in;  //结点i增1次数
    LL de;  //结点i减1次数
}cnt[MAXN];
vector<LL> node[MAXN];

Count DP(int i)
{
    vis[i] = true;
    LL in = 0, de = 0;  //in for increse, de for decrese
    Count t;
    for(vector<LL>::iterator it = node[i].begin(); it != node[i].end(); it++)
    {
        if(vis[*it]) continue;
        t = DP(*it);
        if(t.in > in) in = t.in;
        if(t.de > de) de = t.de;
    }
    cnt[i].de = de;
    cnt[i].in = in;
    int d = v[i] - cnt[i].de + cnt[i].in;
    if(d > 0) cnt[i].de += d;
    else cnt[i].in -= d;
    return cnt[i];
}


int main()
{
    while(scanf("%d", &n) != EOF)
    {
        for(int i = 1; i <= n; i++)
        {
            vis[0] = false;
            while(!node[i].empty()) node[i].pop_back();
        }
        int a, b;
        for(int i = 1; i < n; i++)
        {
            scanf("%d %d", &a, &b);
            node[a].push_back(b);
            node[b].push_back(a);
        }
        for(int i = 1; i <= n; i++)
            scanf("%I64d", &v[i]);
        Count ans = DP(1);
        printf("%I64d\n", ans.de + ans.in);
    }
    return 0;
}