【Codeforces 1037D】Valid BFS?
【链接】 我是链接,点我呀:)
【题意】
【题解】
先dfs出来每一层有多少个点,以及每个点是属于哪一层的。 每一层的bfs如果有先后顺序的话,下一层的节点的出队也是有先后顺序的 因此x是当前层只是一个简单的判断条件,还需要更深入的判断 也就是说它是不是应该在这个时候从队列中出来,也就是说它前面是不是应该有节点比它先出现. 我们需要记录每个节点(dep层)的priority值。 这个值表示了这个节点x它的直系儿子们(dep+1层)在所有dep+1层出现的顺序(如果为1最先出现,其次的话依次往后推 细节比较多。 要注意某个节点可能没有后代节点了。【代码】
import java.io.*;
import java.util.*;
public class Main {
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static int N = (int)2e5;;
static class Task{
int n;
ArrayList g[] = new ArrayList[N+10];
int fa[] = new int[N+10];
int dep[]= new int[N+10],dep_size[] = new int[N+10];
int priority[] = new int[N+10];
void dfs1(int x,int f) {
dep_size[dep[x]]++;
fa[x] = f;
int len = g[x].size();
for (int i = 0;i < len;i++) {
int y = (int)g[x].get(i);
if (y==f) continue;
dep[y] = dep[x] + 1;
dfs1(y,x);
}
}
int get_priority_up(int x) {
if (x==1)
return g[x].size();
else
return g[x].size()-1;
}
public void solve(InputReader in,PrintWriter out) {
for (int i = 1;i <= N;i++) g[i] = new ArrayList<>();
n = in.nextInt();
for (int i = 1;i <= n-1;i++) {
int x,y;
x = in.nextInt();y = in.nextInt();
g[x].add(y);g[y].add(x);
}
dep[1] = 1;
dfs1(1,-1);
int x;
x = in.nextInt();
if (x!=1) {
out.println("No");
return;
}
priority[1] = 1;
int cur_dep = 2,cur_priority = 1,cur_priority_count = 0;
int record_priority = 1;
ArrayList cur_priority_size[] = new ArrayList[N+10];
for (int i = 1;i <= n;i++) {
cur_priority_size[i] = new ArrayList<>();
cur_priority_size[i].add(-1);
}
cur_priority_size[1].add(g[1].size());
for (int i = 2;i <= n;i++) {
while((int)cur_priority_size[cur_dep-1].get(cur_priority)==0) {
cur_priority++;
if (cur_priority>dep_size[cur_dep-1]) {
record_priority = 1;
cur_dep++;
cur_priority = 1;
}
}
x = in.nextInt();
if (dep[x]!=cur_dep) {
out.println("No");
return;
}
int _father = fa[x];
int f_priority = priority[_father];
if (f_priority!=cur_priority) {
out.println("No");
return;
}
priority[x] = record_priority;
cur_priority_size[cur_dep].add(g[x].size()-1);
//out.println("priority["+x+"]="+priority[x]);
record_priority++;
cur_priority_count++;
if (cur_priority_count==get_priority_up(_father)) {
cur_priority++;
cur_priority_count = 0;
if (cur_priority>dep_size[cur_dep-1]) {
record_priority = 1;
cur_dep++;
cur_priority = 1;
}
}
}
out.println("Yes");
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}