CodeForces 321C Ciel the Commander

Ciel the Commander

Time Limit: 1000ms

Memory Limit: 262144KB

This problem will be judged on  CodeForces. Original ID: 321C
64-bit integer IO format: %I64d      Java class name: (Any)

Now Fox Ciel becomes a commander of Tree Land. Tree Land, like its name said, has  n cities connected by n - 1 undirected roads, and for any two cities there always exists a path between them.

 

Fox Ciel needs to assign an officer to each city. Each officer has a rank — a letter from 'A' to 'Z'. So there will be 26 different ranks, and 'A' is the topmost, so 'Z' is the bottommost.

There are enough officers of each rank. But there is a special rule must obey: if x and y are two distinct cities and their officers have the same rank, then on the simple path between xand y there must be a city z that has an officer with higher rank. The rule guarantee that a communications between same rank officers will be monitored by higher rank officer.

Help Ciel to make a valid plan, and if it's impossible, output "Impossible!".

Input

The first line contains an integer n (2 ≤ n ≤ 105) — the number of cities in Tree Land.

Each of the following n - 1 lines contains two integers a and b (1 ≤ a, b ≤ n, a ≠ b) — they mean that there will be an undirected road between a and b. Consider all the cities are numbered from 1 to n.

It guaranteed that the given graph will be a tree.

 

Output

If there is a valid plane, output n space-separated characters in a line — i-th character is the rank of officer in the city with number i.

Otherwise output "Impossible!".

 

Sample Input

Input

4
1 2
1 3
1 4

Output

A B B B

Input

10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10

Output

D C B A D C B D C D

Hint

In the first example, for any two officers of rank 'B', an officer with rank 'A' will be on the path between them. So it is a valid solution.

 

Source

Codeforces Round #190 (Div. 1)

 

解题：树分治

深度大于26，说明impossible

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 100010;
 4 bool done[maxn];
 5 vector<int>g[maxn];
 6 char ans[maxn];
 7 int sz[maxn],maxson[maxn];
 8 int dfs(int u,int fa){
 9     sz[u] = 1;
10     maxson[u] = 0;
11     for(int i = g[u].size()-1; i >= 0; --i){
12         if(g[u][i] == fa || done[g[u][i]]) continue;
13         dfs(g[u][i],u);
14         sz[u] += sz[g[u][i]];
15         maxson[u] = max(maxson[u],sz[g[u][i]]);
16     }
17     return sz[u];
18 }
19 int FindRoot(const int sum,int u,int fa){
20     int ret = u;
21     maxson[u] = max(maxson[u],sum - sz[u]);
22     for(int i = g[u].size()-1; i >= 0; --i){
23         if(g[u][i] == fa || done[g[u][i]]) continue;
24         int x = FindRoot(sum,g[u][i],u);
25         if(maxson[x] < maxson[ret]) ret = x;
26     }
27     return ret;
28 }
29 bool solve(int u,char ch){
30     int root = FindRoot(dfs(u,0),u,0);
31     done[root] = true;
32     ans[root] = ch;
33     if(ch > 'Z') return false;
34     for(int i = g[root].size()-1; i >= 0; --i){
35         if(done[g[root][i]]) continue;
36         if(!solve(g[root][i],ch + 1)) return false;
37     }
38     return true;
39 }
40 int main(){
41     int n,u,v;
42     while(~scanf("%d",&n)){
43         for(int i = 0; i <= n; ++i){
44             done[i] = false;
45             g[i].clear();
46         }
47         for(int i = 1; i < n; ++i){
48             scanf("%d%d",&u,&v);
49             g[u].push_back(v);
50             g[v].push_back(u);
51         }
52         if(solve(1,'A'))
53         for(int i = 1; i <= n; ++i)
54             printf("%c%c",ans[i],i == n?'\n':' ');
55         else puts("Impossible!");
56     }
57     return 0;
58 }

View Code

 

转载于:https://www.cnblogs.com/crackpotisback/p/4950886.html