2022-06-03:a -> b,代表a在食物链中被b捕食, 给定一个有向无环图,返回这个图中从最初级动物到最顶级捕食者的食物链有几条。 来自理想汽车。
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2022-06-03:a -> b,代表a在食物链中被b捕食,
给定一个有向无环图,返回这个图中从最初级动物到最顶级捕食者的食物链有几条。
来自理想汽车。
答案2022-06-03:
拓扑排序。
代码用rust编写。代码如下:
fn main()
let sc: Vec<i32> = vec![5, 7, 1, 2, 1, 3, 2, 3, 3, 5, 2, 5, 4, 5, 3, 4];
let mut ii: i32 = 0;
while ii < sc.len() as i32
let mut info = GlobalInfo::new();
info.n = sc[ii as usize];
ii += 1;
let m: i32 = sc[ii as usize];
ii += 1;
let mut pre_edge: Vec<i32> = vec![];
let mut edges_to: Vec<i32> = vec![];
for _ in 0..m + 1
pre_edge.push(0);
edges_to.push(0);
for i in 1..=m
let from = sc[ii as usize];
ii += 1;
let to = sc[ii as usize];
ii += 1;
edges_to[i as usize] = to;
pre_edge[i as usize] = info.head_edge[from as usize];
info.head_edge[from as usize] = i;
info.out0[from as usize] = true;
info.in0[to as usize] += 1;
let ans = how_many_ways(&mut pre_edge, &mut edges_to, &mut info);
println!("ans = ", ans);
pub struct GlobalInfo
in0: Vec<i32>,
out0: Vec<bool>,
lines: Vec<i32>,
head_edge: Vec<i32>,
queue: Vec<i32>,
mod0: i32,
n: i32,
impl GlobalInfo
pub fn new() -> Self
let mut in0: Vec<i32> = vec![];
let mut out0: Vec<bool> = vec![];
let mut lines: Vec<i32> = vec![];
let mut head_edge: Vec<i32> = vec![];
let mut queue: Vec<i32> = vec![];
let mod0: i32 = 80112002;
let n: i32 = 0;
for _i in 0..5001
in0.push(0);
out0.push(false);
lines.push(0);
head_edge.push(0);
queue.push(0);
Self
in0,
out0,
lines,
head_edge,
queue,
mod0,
n,
fn how_many_ways(pre_edge: &mut Vec<i32>, edges_to: &mut Vec<i32>, info: &mut GlobalInfo) -> i32
let mut ql = 0;
let mut qr = 0;
for i in 1..info.n
if info.in0[i as usize] == 0
info.queue[qr as usize] = i;
qr += 1;
info.lines[i as usize] = 1;
while ql < qr
let cur = info.queue[ql];
ql += 1;
let mut edge = info.head_edge[cur as usize];
while edge != 0
let next = edges_to[edge as usize];
info.lines[next as usize] =
(info.lines[next as usize] + info.lines[cur as usize]) % info.mod0;
info.in0[next as usize] -= 1;
if info.in0[next as usize] == 0
info.queue[qr] = next;
qr += 1;
edge = pre_edge[edge as usize];
let mut ans = 0;
for i in 1..=info.n
if !info.out0[i as usize]
ans = (ans + info.lines[i as usize]) % info.mod0;
return ans;
执行结果如下:
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