problem
C. Parsa’s Humongous Tree
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Parsa has a humongous tree on n vertices.
On each vertex v he has written two integers lv and rv.
To make Parsa’s tree look even more majestic, Nima wants to assign a number av (lv≤av≤rv) to each vertex v such that the beauty of Parsa’s tree is maximized.
Nima’s sense of the beauty is rather bizarre. He defines the beauty of the tree as the sum of |au−av| over all edges (u,v) of the tree.
Since Parsa’s tree is too large, Nima can’t maximize its beauty on his own. Your task is to find the maximum possible beauty for Parsa’s tree.
Input
The first line contains an integer t (1≤t≤250) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2≤n≤105) — the number of vertices in Parsa’s tree.
The i-th of the following n lines contains two integers li and ri (1≤li≤ri≤109).
Each of the next n−1 lines contains two integers u and v (1≤u,v≤n,u≠v) meaning that there is an edge between the vertices u and v in Parsa’s tree.
It is guaranteed that the given graph is a tree.
It is guaranteed that the sum of n over all test cases doesn’t exceed 2⋅105.
Output
For each test case print the maximum possible beauty for Parsa’s tree.
Example
inputCopy
3
2
1 6
3 8
1 2
3
1 3
4 6
7 9
1 2
2 3
6
3 14
12 20
12 19
2 12
10 17
3 17
3 2
6 5
1 5
2 6
4 6
outputCopy
7
8
62
Note
The trees in the example:
In the first test case, one possible assignment is a={1,8} which results in |1−8|=7.
In the second test case, one of the possible assignments is a={1,5,9} which results in a beauty of |1−5|+|5−9|=8
solution
//求相邻点的最大绝对值之差,那么肯定是取两个点的左端点和右端点最优
//f[x][0]表示ax=l时，x的子树的美丽值之和,f[x][1]表示ax=r
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e6+10;
vector<int>G[maxn];
LL l[maxn], r[maxn], f[maxn][2];
void dfs(int u, int fa){
for(int v: G[u]){
if(v==fa)continue;
dfs(v,u);
f[u][0] += max(f[v][0]+abs(l[u]-l[v]), f[v][1]+abs(l[u]-r[v]));
f[u][1] += max(f[v][0]+abs(r[u]-l[v]), f[v][1]+abs(r[u]-r[v]));
}
}
int main(){
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int T; cin>>T;
while(T--){
int n; cin>>n;
for(int i = 1; i <= n; i++){
cin>>l[i]>>r[i];
f[i][0] = f[i][1] = 0;
G[i].clear();
}
for(int i = 1; i <= n-1; i++){
int u, v; cin>>u>>v;
G[u].push_back(v);
G[v].push_back(u);
}
dfs(1,0);
LL ans = 0;
for(int i = 1; i <= n; i++)
ans = max(ans, max(f[i][0],f[i][1]) );
cout<<ans<<"\n";
}
return 0;
}