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:

Select the subtree of the given tree that includes the vertex with number 1.

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

一開始題目理解錯誤：“includes the vertex with number 1.”是指結(jié)點1而非其上integer為1的結(jié)點。
思路是先處理兒子再處理父親。這是因為：設(shè)有結(jié)點x，x的val變?yōu)?所需操作次數(shù)x.count = in + de + r,這里in是x因為兒子結(jié)點的+1操作
而隨之產(chǎn)生的+1操作的次數(shù)（對兒子的操作會傳遞到父親），而de就是-1操作，顯然，in是x的所有兒子中+1次數(shù)的最大值，de是所有兒子中-1次
數(shù)的最大值。而r是什么？在處理完兒子后確保兒子的val為0，但不保證x的val為0，r就是余下的對x操作的次數(shù)：r=x.val+in-de（這些操作是
發(fā)生在x和他的父親之間的）。由以上于可見，對x.count的統(tǒng)計能轉(zhuǎn)化為對其兒子結(jié)點的相關(guān)信息的統(tǒng)計，故而此前說“先處理兒子后處理父親”。
結(jié)點1是最后處理的。
這里還涉及如何保存一個結(jié)點的所有兒子的信息的問題，我開始用鄰接表，超時，后來參考別人的代碼才想起來可以用vector。

AC Code：
1 #include <iostream> 2 #include <string> 3 #include <set> 4 #include <map> 5 #include <vector> 6 #include <stack> 7 #include <queue> 8 #include <cmath> 9 #include <cstdio> 10 #include <cstring> 11 #include <algorithm> 12 using namespace std; 13 #define LL long long 14 #define cti const int 15 #define dg(i) cout << "*" << i << endl; 16 17 cti MAXN = 100005; 18 int n; 19 bool vis[MAXN]; 20 LL v[MAXN]; 21 struct Count 22 { 23 LL in; //結(jié)點i增1次數(shù) 24 LL de; //結(jié)點i減1次數(shù) 25 }cnt[MAXN]; 26 vector<LL> node[MAXN]; 27 28 Count DP(int i) 29 { 30 vis[i] = true; 31 LL in = 0, de = 0; //in for increse, de for decrese 32 Count t; 33 for(vector<LL>::iterator it = node[i].begin(); it != node[i].end(); it++) 34 { 35 if(vis[*it]) continue; 36 t = DP(*it); 37 if(t.in > in) in = t.in; 38 if(t.de > de) de = t.de; 39 } 40 cnt[i].de = de; 41 cnt[i].in = in; 42 int d = v[i] - cnt[i].de + cnt[i].in; 43 if(d > 0) cnt[i].de += d; 44 else cnt[i].in -= d; 45 return cnt[i]; 46 } 47 48 49 int main() 50 { 51 while(scanf("%d", &n) != EOF) 52 { 53 for(int i = 1; i <= n; i++) 54 { 55 vis[0] = false; 56 while(!node[i].empty()) node[i].pop_back(); 57 } 58 int a, b; 59 for(int i = 1; i < n; i++) 60 { 61 scanf("%d %d", &a, &b); 62 node[a].push_back(b); 63 node[b].push_back(a); 64 } 65 for(int i = 1; i <= n; i++) 66 scanf("%I64d", &v[i]); 67 Count ans = DP(1); 68 printf("%I64d\n", ans.de + ans.in); 69 } 70 return 0; 71 }

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