A843.Balancing a Tree--Gold

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时间限制:1.00s

内存限制:128MB

题目描述

Farmer John has conducted an extensive study of the evolution of different cow
breeds. The result is a rooted tree with NN (2≤N≤1052\le N\le 10^5) nodes labeled
1…N1\ldots N, each node corresponding to a cow breed. For each i∈[2,N]i\in [2,N],
the parent of node ii is node pip_i (1≤pi<i1\le p_i<i), meaning that breed ii
evolved from breed pip_i. A node jj is called an ancestor of node ii if
j=pij=p_i or jj is an ancestor of pip_i.
Every node ii in the tree is associated with a breed having an integer number
of spots sis_i. The "imbalance" of the tree is defined to be the maximum of
∣si−sj∣|s_i-s_j| over all pairs of nodes (i,j)(i,j) such that jj is an ancestor of
ii.
Farmer John doesn't know the exact value of sis_i for each breed, but he knows
lower and upper bounds on these values. Your job is to assign an integer value
of si∈[li,ri]s_i \in [l_i,r_i] (0≤li≤ri≤1090\le l_i\le r_i\le 10^9) to each node such that the
imbalance of the tree is minimized.

输入格式

The first line contains TT (1≤T≤101\le T\le 10), the number of independent test
cases to be solved, and an integer B∈0,1B\in \\{0,1\\}.
Each test case starts with a line containing NN, followed by N−1N-1 integers
p2,p3,…,pNp_2,p_3,\ldots,p_N.
The next NN lines each contain two integers lil_i and rir_i.
It is guaranteed that the sum of NN over all test cases does not exceed
10510^5.

输出格式

For each test case, output one or two lines, depending on the value of BB.
The first line for each test case should contain the minimum imbalance.
If B=1,B=1, then print an additional line with NN space-separated integers
s1,s2,…,sNs_1,s_2,\ldots, s_N containing an assignment of spots that achieves the
above imbalance. Any valid assignment will be accepted.

输入输出样例

  • 输入#1

    3 0
    3
    1 1
    0 100
    1 1
    6 7
    5
    1 2 3 4
    6 6
    1 6
    1 6
    1 6
    5 5
    3
    1 1
    0 10
    0 1
    9 10
    

    输出#1

    3
    1
    4
    

说明/提示

For the first test case, the minimum imbalance is 33. One way to achieve
imbalance 33 is to set [s1,s2,s3]=[4,1,7][s_1,s_2,s_3]=[4,1,7].

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