結果
問題 | No.1812 Uribo Road |
ユーザー | koba-e964 |
提出日時 | 2022-02-17 22:59:24 |
言語 | Rust (1.77.0 + proconio) |
結果 |
AC
|
実行時間 | 76 ms / 5,000 ms |
コード長 | 3,592 bytes |
コンパイル時間 | 13,370 ms |
コンパイル使用メモリ | 377,972 KB |
実行使用メモリ | 6,944 KB |
最終ジャッジ日時 | 2024-06-29 07:52:52 |
合計ジャッジ時間 | 14,314 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge4 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 1 ms
6,816 KB |
testcase_01 | AC | 1 ms
6,944 KB |
testcase_02 | AC | 1 ms
6,940 KB |
testcase_03 | AC | 7 ms
6,944 KB |
testcase_04 | AC | 1 ms
6,940 KB |
testcase_05 | AC | 1 ms
6,940 KB |
testcase_06 | AC | 1 ms
6,944 KB |
testcase_07 | AC | 1 ms
6,940 KB |
testcase_08 | AC | 2 ms
6,940 KB |
testcase_09 | AC | 3 ms
6,944 KB |
testcase_10 | AC | 2 ms
6,940 KB |
testcase_11 | AC | 2 ms
6,940 KB |
testcase_12 | AC | 40 ms
6,940 KB |
testcase_13 | AC | 20 ms
6,940 KB |
testcase_14 | AC | 20 ms
6,944 KB |
testcase_15 | AC | 24 ms
6,940 KB |
testcase_16 | AC | 25 ms
6,940 KB |
testcase_17 | AC | 61 ms
6,944 KB |
testcase_18 | AC | 16 ms
6,944 KB |
testcase_19 | AC | 75 ms
6,940 KB |
testcase_20 | AC | 76 ms
6,944 KB |
testcase_21 | AC | 62 ms
6,940 KB |
testcase_22 | AC | 63 ms
6,944 KB |
testcase_23 | AC | 3 ms
6,940 KB |
testcase_24 | AC | 1 ms
6,944 KB |
testcase_25 | AC | 10 ms
6,944 KB |
testcase_26 | AC | 1 ms
6,940 KB |
testcase_27 | AC | 11 ms
6,944 KB |
testcase_28 | AC | 19 ms
6,944 KB |
testcase_29 | AC | 1 ms
6,944 KB |
testcase_30 | AC | 3 ms
6,940 KB |
testcase_31 | AC | 38 ms
6,944 KB |
testcase_32 | AC | 1 ms
6,940 KB |
testcase_33 | AC | 30 ms
6,940 KB |
ソースコード
use std::cmp::*; // https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 macro_rules! input { ($($r:tt)*) => { let stdin = std::io::stdin(); let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock())); let mut next = move || -> String{ bytes.by_ref().map(|r|r.unwrap() as char) .skip_while(|c|c.is_whitespace()) .take_while(|c|!c.is_whitespace()) .collect() }; input_inner!{next, $($r)*} }; } macro_rules! input_inner { ($next:expr) => {}; ($next:expr,) => {}; ($next:expr, $var:ident : $t:tt $($r:tt)*) => { let $var = read_value!($next, $t); input_inner!{$next $($r)*} }; } macro_rules! read_value { ($next:expr, ( $($t:tt),* )) => { ($(read_value!($next, $t)),*) }; ($next:expr, [ $t:tt ; $len:expr ]) => { (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>() }; ($next:expr, usize1) => (read_value!($next, usize) - 1); ($next:expr, $t:ty) => ($next().parse::<$t>().expect("Parse error")); } /* * Dijkstra's algorithm. * Verified by: AtCoder ABC164 (https://atcoder.jp/contests/abc164/submissions/12423853) */ struct Dijkstra { edges: Vec<Vec<(usize, i64)>>, // adjacent list representation } impl Dijkstra { fn new(n: usize) -> Self { Dijkstra { edges: vec![Vec::new(); n] } } fn add_edge(&mut self, from: usize, to: usize, cost: i64) { self.edges[from].push((to, cost)); } /* * This function returns a Vec consisting of the distances from vertex source. */ fn solve(&self, source: usize, inf: i64) -> Vec<i64> { let n = self.edges.len(); let mut d = vec![inf; n]; // que holds (-distance, vertex), so that que.pop() returns the nearest element. let mut que = std::collections::BinaryHeap::new(); que.push((0, source)); while let Some((cost, pos)) = que.pop() { let cost = -cost; if d[pos] <= cost { continue; } d[pos] = cost; for &(w, c) in &self.edges[pos] { let newcost = cost + c; if d[w] > newcost { d[w] = newcost + 1; que.push((-newcost, w)); } } } return d; } } fn main() { input! { n: usize, m: usize, k: usize, r: [usize1; k], abc: [(usize1, usize1, i64); m], } let mut dijk = Dijkstra::new(n); for &(a, b, c) in &abc { dijk.add_edge(a, b, c); dijk.add_edge(b, a, c); } const INF: i64 = 1 << 50; let s0 = dijk.solve(0, INF); let s1 = dijk.solve(n - 1, INF); let mut s = vec![vec![]; 2 * k]; let mut v = vec![0; 2 * k]; for i in 0..k { let (a, b, _) = abc[r[i]]; s[2 * i] = dijk.solve(a, INF); s[2 * i + 1] = dijk.solve(b, INF); v[2 * i] = a; v[2 * i + 1] = b; } let mut dp = vec![vec![INF; 1 << k]; 2 * k]; for i in 0..2 * k { dp[i ^ 1][1 << (i / 2)] = s0[v[i]] + abc[r[i / 2]].2; } for bits in 1..1 << k { for j in 0..2 * k { if (bits & 1 << (j / 2)) == 0 { continue; } let pre = bits ^ 1 << (j / 2); for l in 0..2 * k { dp[j][bits] = min(dp[j][bits], dp[l][pre] + s[l][v[j ^ 1]] + abc[r[j / 2]].2); } } } let mut ans = INF; for j in 0..2 * k { ans = min(ans, dp[j][(1 << k) - 1] + s1[v[j]]); } println!("{}", ans); }