Problem:
Score : 700 points

Problem Statement

We have a tree with N vertices. Vertex 1 is the root of the tree, and the parent of Vertex i (2 \leq i \leq N) is Vertex P_i.

To each vertex in the tree, Snuke will allocate a color, either black or white, and a non-negative integer weight.

Snuke has a favorite integer sequence, X_1, X_2, ..., X_N, so he wants to allocate colors and weights so that the following condition is satisfied for all v.

The total weight of the vertices with the same color as v among the vertices contained in the subtree whose root is v, is X_v.

Here, the subtree whose root is v is the tree consisting of Vertex v and all of its descendants.

Determine whether it is possible to allocate colors and weights in this way.

Constraints

1 \leq N \leq 1 000

1 \leq P_i \leq i - 1

0 \leq X_i \leq 5 000

Inputs

Input is given from Standard Input in the following format:

N
P_2 P_3 ... P_N
X_1 X_2 ... X_N

Outputs

If it is possible to allocate colors and weights to the vertices so that the condition is satisfied, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

3
1 1
4 3 2

Sample Output 1

POSSIBLE

For example, the following allocation satisfies the condition:

Set the color of Vertex 1 to white and its weight to 2.

Set the color of Vertex 2 to black and its weight to 3.

Set the color of Vertex 3 to white and its weight to 2.

There are also other possible allocations.

Sample Input 2

3
1 2
1 2 3

Sample Output 2

IMPOSSIBLE

If the same color is allocated to Vertex 2 and Vertex 3, Vertex 2 cannot be allocated a non-negative weight.

If different colors are allocated to Vertex 2 and 3, no matter which color is allocated to Vertex 1, it cannot be allocated a non-negative weight.

Thus, there exists no allocation of colors and weights that satisfies the condition.

Sample Input 3

8
1 1 1 3 4 5 5
4 1 6 2 2 1 3 3

Sample Output 3

POSSIBLE

Sample Input 4

1

0

Sample Output 4

POSSIBLE

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

#![allow(unused_imports, unused_variables, dead_code)]
use std::io::*;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

trait InputValue {
    fn parse(s: &str) -> Self;
}

fn read<T: InputValue>() -> T {
    let mut buf = String::new();
    let _ = stdin().read_line(&mut buf);
    T::parse(&buf.trim())
}

fn readnc<T: InputValue>() -> Vec<T> {
    let mut vec = vec![];
    let line: String = read();
    for token in line.split_whitespace() {
        vec.push(T::parse(token));
    }
    vec
}

fn readn<T: InputValue>(n: usize) -> Vec<T> {
    let mut vec = vec![];
    for _ in 0..n {
        vec.push(read());
    }
    vec
}

macro_rules! parse_single_value {
    ($($t:ty),*) => {
        $(
            impl InputValue for $t {
                fn parse(s: &str) -> $t { s.parse().unwrap() }
            }
        )*
	}
}
parse_single_value!(i32, i64, f32, f64, usize, String);

macro_rules! parse_tuple {
	($($t:ident),*) => {
		impl<$($t),*> InputValue for ($($t),*) where $($t: InputValue),* {
			fn parse(s: &str) -> ($($t),*) {
				let mut tokens = s.split_whitespace();
				let t = ($($t::parse(tokens.next().unwrap())),*);
				t
			}
		}
	}
}
parse_tuple!(A, B);
parse_tuple!(A, B, C);

// ===

const INF: usize = 10000;
const MAX: usize = 5010;

fn main() {
    let n: usize = read();
    let mut p: Vec<usize> = readnc();
    p.insert(0, n);
    for i in 1..n {
        p[i] -= 1;
    }

    let x: Vec<usize> = readnc();

    let mut children: Vec<Vec<usize>> = vec![vec![]; n];
    for i in 1..n {
        children[p[i]].push(i);
    }


    let mut best: Vec<usize> = vec![INF; n];
    let mut dp: Vec<Vec<usize>> = vec![vec![INF; MAX]; 2];



    let mut isok = true;
    for i in (0..n).rev() {
        let cn = children[i].len();
        for i in 0..MAX {
            dp[0][i] = INF;
            dp[1][i] = INF;
        }

        dp[0][0] = 0;
        for ci in 0..cn {
            let a = x[children[i][ci]];
            let b = best[children[i][ci]];

            let fi = ci % 2;
            let ti = 1 - fi;
            for i in 0..MAX {
                dp[ti][i] = INF;
            }
            for i in 0..MAX {
                if dp[fi][i] == INF {
                    continue
                }
                let base = dp[fi][i];
                let ia = i + a;
                if ia < MAX {
                    dp[ti][ia] = min(dp[ti][ia], base + b);
                }
                let ib = i + b;
                if ib < MAX {
                    dp[ti][ib] = min(dp[ti][ib], base + a);
                }
            }
        }

        let wi = cn % 2;
        let mut best_op = INF;
        for l in 0..x[i]+1 {
            best_op = min(best_op, dp[wi][l]);
        }

        if best_op == INF {
            isok = false;
            break
        }
        best[i] = best_op;
    }

    if isok {
        println!("POSSIBLE");
    } else {
        println!("IMPOSSIBLE");
    }
}