Problem:
Score : 400 points

Problem Statement

Kenkoooo is planning a trip in Republic of Snuke.
In this country, there are n cities and m trains running.
The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally.
Any city can be reached from any city by changing trains.

Two currencies are used in the country: yen and snuuk.
Any train fare can be paid by both yen and snuuk.
The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.

In a city with a money exchange office, you can change 1 yen into 1 snuuk.
However, when you do a money exchange, you have to change all your yen into snuuk.
That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk.
Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.

Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling.
It is acceptable to do the exchange in City s or City t.

Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange.
For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years.
You can assume that the trip finishes within the year.

Constraints

2 \leq n \leq 10^5

1 \leq m \leq 10^5

1 \leq s,t \leq n

s \neq t

1 \leq u_i < v_i \leq n

1 \leq a_i,b_i \leq 10^9

If i\neq j, then u_i \neq u_j or v_i \neq v_j.

Any city can be reached from any city by changing trains.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

n m s t
u_1 v_1 a_1 b_1
:
u_m v_m a_m b_m

Output

Print n lines.
In the i-th line, print the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i-1 years.

Sample Input 1

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1

Sample Output 1

999999999999998
999999999999989
999999999999979
999999999999897

After 0 years, it is optimal to do the exchange in City 1.

After 1 years, it is optimal to do the exchange in City 2.

Note that City 1 can still be visited even after the exchange office is closed.
Also note that, if it was allowed to keep 1 yen when do the exchange in City 2 and change the remaining yen into snuuk, we could reach City 3 with 999999999999998 snuuk, but this is NOT allowed.

After 2 years, it is optimal to do the exchange in City 3.

After 3 years, it is optimal to do the exchange in City 4.
Note that the same train can be used multiple times.

Sample Input 2

8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879

Sample Output 2

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

Solution in Rust (1.15.1) (judge verdict: Time Limit Exceeded)

//  ____            _                  _                       _       _
// |  _ \ _   _ ___| |_    ___ _ __   | |_ ___ _ __ ___  _ __ | | __ _| |_ ___
// | |_) | | | / __| __|  / __| '_ \  | __/ _ \ '_ ` _ \| '_ \| |/ _` | __/ _ \
// |  _ <| |_| \__ \ |_  | (__| |_) | | ||  __/ | | | | | |_) | | (_| | ||  __/
// |_| \_\\__,_|___/\__|  \___| .__/___\__\___|_| |_| |_| .__/|_|\__,_|\__\___|
//                            |_| |_____|               |_|

//                                                     _        _ _ _____  ___
//                               _ __ ___   __ _ _ __ | |_ __ _/ / |___ / / _ \
//                              | '_ ` _ \ / _` | '_ \| __/ _` | | | |_ \| | | |
//                              | | | | | | (_| | | | | || (_| | | |___) | |_| |
//                              |_| |_| |_|\__,_|_| |_|\__\__,_|_|_|____/ \___/

//https://github.com/manta1130/Competitive_Programming_Template_Rust

#[macro_use]
mod input {

    use std;
    use std::io;

    const SPLIT_DELIMITER: char = ' ';

    #[macro_export]
    #[allow(unused_macros)]
    macro_rules! input {
    ( $($x:expr ),*) => {
        {
            let temp_str = input_line_str();
            let mut split_result_iter = temp_str.split_whitespace();
                $(
                let buf_split_result = split_result_iter.next();
                let buf_split_result = buf_split_result.unwrap();
                    ($x) = buf_split_result.parse().unwrap();
                )*
        }
    };
    }

    #[allow(dead_code)]
    pub fn input_line_str() -> String {
        let mut s = String::new();
        io::stdin().read_line(&mut s).unwrap();
        s.trim().to_string()
    }

    #[allow(dead_code)]
    pub fn p<T>(t: T)
    where
        T: std::fmt::Display,
    {
        println!("{}", t);
    }

    #[allow(dead_code)]
    pub fn input_vector2d<T>(line: usize) -> Vec<Vec<T>>
    where
        T: std::str::FromStr,
    {
        let mut v: Vec<Vec<T>> = Vec::new();

        for _ in 0..line {
            let vec_line = input_vector();
            v.push(vec_line);
        }
        v
    }

    #[allow(dead_code)]
    pub fn input_vector<T>() -> Vec<T>
    where
        T: std::str::FromStr,
    {
        let mut v: Vec<T> = Vec::new();

        let s = input_line_str();
        let split_result = s.split(SPLIT_DELIMITER);
        for z in split_result {
            let buf = match z.parse() {
                Ok(r) => r,
                Err(_) => panic!("Parse Error"),
            };
            v.push(buf);
        }
        v
    }

    #[allow(dead_code)]
    pub fn input_vector_row<T>(n: usize) -> Vec<T>
    where
        T: std::str::FromStr,
    {
        let mut v = Vec::with_capacity(n);
        for _ in 0..n {
            let buf = match input_line_str().parse() {
                Ok(r) => r,
                Err(_) => panic!("Parse Error"),
            };
            v.push(buf);
        }
        v
    }

    pub trait ToCharVec {
        fn to_charvec(&self) -> Vec<char>;
    }

    impl ToCharVec for String {
        fn to_charvec(&self) -> Vec<char> {
            self.to_string().chars().collect::<Vec<_>>()
        }
    }
}

use input::*;
use std::cmp::{Ord, Ordering, PartialOrd};
use std::collections::BinaryHeap;

#[derive(Eq, PartialEq)]
struct Edge {
    cost: isize,
    vertex: usize,
}

impl Ord for Edge {
    fn cmp(&self, other: &Self) -> Ordering {
        other.cost.cmp(&self.cost)
    }
}
impl PartialOrd for Edge {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

#[derive(PartialEq, Eq, Clone)]
struct Train {
    to: usize,
    cost_yen: isize,
    cost_snuke: isize,
}

const INF: isize = 999_999_999_999_999;

fn main() {
    let (n, m, mut s, mut t): (usize, usize, usize, usize);
    input!(n, m, s, t);
    let mut graph = vec![vec![]; n];
    let mut shortest_yen = vec![INF; n];
    let mut shortest_snuke = vec![INF; n];
    let mut heap = BinaryHeap::new();

    s -= 1;
    t -= 1;

    shortest_yen[s] = 0;
    shortest_snuke[t] = 0;

    for _ in 0..m {
        let (u, v, a, b): (usize, usize, isize, isize);
        input!(u, v, a, b);
        graph[u - 1].push(Train {
            to: v - 1,
            cost_yen: a,
            cost_snuke: b,
        });
        graph[v - 1].push(Train {
            to: u - 1,
            cost_yen: a,
            cost_snuke: b,
        });
    }

    for e in &graph[s] {
        heap.push(Edge {
            cost: e.cost_yen,
            vertex: e.to,
        });
    }

    while !heap.is_empty() {
        let e = heap.pop().unwrap();
        if shortest_yen[e.vertex] < e.cost {
            continue;
        }
        shortest_yen[e.vertex] = e.cost;
        for ed in &graph[e.vertex] {
            heap.push(Edge {
                cost: e.cost + ed.cost_yen,
                vertex: ed.to,
            });
        }
    }

    heap.clear();

    for e in &graph[t] {
        heap.push(Edge {
            cost: e.cost_snuke,
            vertex: e.to,
        });
    }

    while !heap.is_empty() {
        let e = heap.pop().unwrap();
        if shortest_snuke[e.vertex] < e.cost {
            continue;
        }
        shortest_snuke[e.vertex] = e.cost;
        for ed in &graph[e.vertex] {
            heap.push(Edge {
                cost: e.cost + ed.cost_snuke,
                vertex: ed.to,
            });
        }
    }

    let mut ans = BinaryHeap::new();
    for k in 0..n {
        ans.push((-1 * (shortest_yen[k] + shortest_snuke[k]), k));
    }

    for k in 0..n {
        let mut pop_buf = *ans.peek().unwrap();
        while pop_buf.1 < k {
            ans.pop();
            pop_buf = *ans.peek().unwrap();
        }
        p(1_000_000_000_000_000 + pop_buf.0);
    }
}