2023-02-14:魔物了占领若干据点,这些据点被若干条道路相连接, roads[i] = [x, y] 表示编号 xy 的两个据点通过一条道路连接。 现在勇者要将按照以下原则将这些据点逐一夺回:
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2023-02-14:魔物了占领若干据点,这些据点被若干条道路相连接,
roads[i] = [x, y] 表示编号 x、y 的两个据点通过一条道路连接。
现在勇者要将按照以下原则将这些据点逐一夺回:
在开始的时候,勇者可以花费资源先夺回一些据点,
初始夺回第 j 个据点所需消耗的资源数量为 cost[j]
接下来,勇者在不消耗资源情况下,
每次可以夺回一个和「已夺回据点」相连接的魔物据点,
并对其进行夺回。
为了防止魔物暴动,勇者在每一次夺回据点后(包括花费资源夺回据点后),
需要保证剩余的所有魔物据点之间是相连通的(不经过「已夺回据点」)。
请返回勇者夺回所有据点需要消耗的最少资源数量。
输入保证初始所有据点都是连通的,且不存在重边和自环。
输入:cost = [1,2,3,4,5,6],roads = [[0,1],[0,2],[1,3],[2,3],[1,2],[2,4],[2,5]]。
输出:6。
答案2023-02-24:
代码用rust编写。代码如下:
执行结果如下:
use std::iter::repeat;
fn main()
let cost = vec![1, 2, 3, 4, 5, 6];
let roads = vec![
vec![0, 1],
vec![0, 2],
vec![1, 3],
vec![2, 3],
vec![1, 2],
vec![2, 4],
vec![2, 5],
];
let ans = unsafe Solution::minimum_cost(cost, roads) ;
println!("ans = :?", ans);
struct Solution
impl Solution
pub fn minimum_cost(cost: Vec<i32>, roads: Vec<Vec<i32>>) -> i64
let mut roads = roads;
let n = cost.len() as i32;
if n == 1
return cost[0] as i64;
let m = roads.len() as i32;
let mut dc = DoubleConnectedComponents::new(n, m, &mut roads);
let mut ans: i64 = 0;
// dcc a,b,c c,d,e
if dc.dcc.len() == 1
ans = i64::MAX;
for num in cost.iter()
ans = get_min(ans, *num as i64);
else
// 不只一个点双连通分量
let mut arr: Vec<i32> = vec![];
for set in dc.dcc.iter()
let mut cutCnt = 0;
let mut curCost = i32::MAX;
for nodes in set.iter()
if dc.cut[*nodes as usize]
cutCnt += 1;
else
curCost = get_min(curCost, cost[*nodes as usize]);
if cutCnt == 1
arr.push(curCost);
arr.sort_by(|a, b| a.partial_cmp(b).unwrap());
for i in 0..arr.len() as i32 - 1
ans += arr[i as usize] as i64;
return ans;
fn get_min<T: Clone + Copy + std::cmp::PartialOrd>(a: T, b: T) -> T
if a < b
a
else
b
struct DoubleConnectedComponents
// 链式前向星建图
head: Vec<i32>,
next: Vec<i32>,
to: Vec<i32>,
dfn: Vec<i32>,
low: Vec<i32>,
stack: Vec<i32>,
dcc: Vec<Vec<i32>>,
cut: Vec<bool>,
edgeCnt: i32,
dfnCnt: i32,
top: i32,
root: i32,
impl DoubleConnectedComponents
pub fn new(n: i32, m: i32, roads: &mut Vec<Vec<i32>>) -> Self
let mut ans = DoubleConnectedComponents
head: vec![],
next: vec![],
to: vec![],
dfn: vec![],
low: vec![],
stack: vec![],
dcc: vec![],
cut: vec![],
edgeCnt: 0,
dfnCnt: 0,
top: 0,
root: 0,
;
ans.init(n, m);
ans.createGraph(roads);
ans.creatDcc(n);
ans
fn init(&mut self, n: i32, m: i32)
let t = repeat(-1).take(n as usize).collect::<Vec<i32>>();
self.head = t;
let t = repeat(0).take((m << 1) as usize).collect::<Vec<i32>>();
self.next = t;
let t = repeat(0).take((m << 1) as usize).collect::<Vec<i32>>();
self.to = t;
let t = repeat(0).take(n as usize).collect::<Vec<i32>>();
self.dfn = t;
let t = repeat(0).take(n as usize).collect::<Vec<i32>>();
self.low = t;
let t = repeat(0).take(n as usize).collect::<Vec<i32>>();
self.stack = t;
let t = repeat(false).take(n as usize).collect::<Vec<bool>>();
self.cut = t;
self.edgeCnt = 0;
self.dfnCnt = 0;
self.top = 0;
self.root = 0;
fn createGraph(&mut self, roads: &mut Vec<Vec<i32>>)
for edges in roads.iter()
self.add(edges[0], edges[1]);
self.add(edges[1], edges[0]);
fn add(&mut self, u: i32, v: i32)
self.to[self.edgeCnt as usize] = v;
self.next[self.edgeCnt as usize] = self.head[u as usize];
self.head[u as usize] = self.edgeCnt;
self.edgeCnt += 1;
fn creatDcc(&mut self, n: i32)
for i in 0..n
// 0 1 2 3 n-1
if self.dfn[i as usize] == 0
self.root = i;
self.tarjan(i);
fn tarjan(&mut self, x: i32)
self.dfnCnt += 1;
self.low[x as usize] = self.dfnCnt;
self.dfn[x as usize] = self.low[x as usize];
self.stack[self.top as usize] = x;
self.top += 1;
let mut flag = 0;
if x == self.root && self.head[x as usize] == -1
self.dcc.push(vec![]);
let t = (self.dcc.len() as i32 - 1) as usize;
self.dcc[t].push(x);
else
// 当前来到的节点是x
// x a,b,c
let mut i = self.head[x as usize];
while i >= 0
// y是下级节点
let mut y = self.to[i as usize];
if self.dfn[y as usize] == 0
// y点没遍历过!
self.tarjan(y);
if self.low[y as usize] >= self.dfn[x as usize]
// 正在扎口袋
flag += 1;
if x != self.root || flag > 1
self.cut[x as usize] = true;
let mut curAns: Vec<i32> = vec![];
// 从栈里一次弹出节点
// 弹到y停!
// 弹出的节点都加入集合,x也加入,x不弹出
self.top -= 1;
let mut z = self.stack[self.top as usize];
while z != y
curAns.push(z);
self.top -= 1;
z = self.stack[self.top as usize];
curAns.push(y);
curAns.push(x);
self.dcc.push(curAns);
self.low[x as usize] = get_min(self.low[x as usize], self.low[y as usize]);
else
// y点已经遍历过了!
self.low[x as usize] = get_min(self.low[x as usize], self.dfn[y as usize]);
i = self.next[i as usize];
以上是关于2023-02-14:魔物了占领若干据点,这些据点被若干条道路相连接, roads[i] = [x, y] 表示编号 xy 的两个据点通过一条道路连接。 现在勇者要将按照以下原则将这些据点逐一夺回:的主要内容,如果未能解决你的问题,请参考以下文章
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