Codeforces 321C Ciel the Commander

Ciel the Commandertime limit:1000msmemory limit:262144kbthis problem would be judged onCodeforces. Original id:321c
64-bit integer IO format: %i64d Java class name: (any) Now Fox Ciel becomes a commander of the Tree land. Tree land, like its name said, have N cities connected by N -1 undirected roads, and for any of the cities there always exists a path Between them.

Fox Ciel needs to assign a officer to each city. Each officer have a rank-a letter from 'a ' to 'Z '. So there would be different ranks, and 'A ' are the topmost, so ' Z ' is thebottommost.

There is enough officers of each rank. But there was a special rule must obey:if x and y are both distinct cities and their officers hav E the same rank, then on the simple path between xand y There must is a city z that have an officer with higher rank. The rule guarantee that a communications between same rank officers would 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 a integer n (2≤ n ≤105)-the number of cities in Tree land.

Each of the following n -1 lines contains double integers a and b (1≤ a, b ≤ n, a ≠ b)-they mean that there'll be is an undirected road BETW Een a and b. Consider all the cities is numbered from 1 to n.

It guaranteed that the given graph would 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 InputInput

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 and officers of rank 'B ', an officer with rank ' A ' would be on thepath between them. So it's a valid solution.

SourceCodeforces Round #190 (Div. 1) Solve the problem: the depth of tree division is greater than 26, indicating impossible

1#include <bits/stdc++.h>2 using namespacestd;3 Const intMAXN =100010;4 BOOLDONE[MAXN];5vector<int>G[MAXN];6 CharANS[MAXN];7 intSZ[MAXN],MAXSON[MAXN];8 intDfsintUintFA) {9Sz[u] =1;TenMaxson[u] =0; One      for(inti = g[u].size ()-1; I >=0; --i) { A         if(G[u][i] = = FA | | done[g[u][i]])Continue; - DFS (g[u][i],u); -Sz[u] + =Sz[g[u][i]]; theMaxson[u] =Max (Maxson[u],sz[g[u][i]]); -     } -     returnSz[u]; - } + intFindRoot (Const intSumintUintFA) { -     intRET =u; +Maxson[u] = max (Maxson[u],sum-Sz[u]); A      for(inti = g[u].size ()-1; I >=0; --i) { at         if(G[u][i] = = FA | | done[g[u][i]])Continue; -         intx =FindRoot (sum,g[u][i],u); -         if(Maxson[x] < Maxson[ret]) ret =x; -     } -     returnret; - } in BOOLSolveintUCharch) { -     introot = FindRoot (Dfs (U,0), U,0); toDone[root] =true; +Ans[root] =ch; -     if(Ch >'Z')return false; the      for(inti = g[root].size ()-1; I >=0; --i) { *         if(Done[g[root][i]])Continue; $         if(!solve (G[root][i],ch +1))return false;Panax Notoginseng     } -     return true; the } + intMain () { A     intn,u,v; the      while(~SCANF ("%d",&N)) { +          for(inti =0; I <= N; ++i) { -Done[i] =false; $ g[i].clear (); $         } -          for(inti =1; I < n; ++i) { -scanf"%d%d",&u,&v); the G[u].push_back (v); - g[v].push_back (u);Wuyi         } the         if(Solve (1,'A')) -          for(inti =1; I <= N; ++i) Wuprintf"%c%c", ans[i],i = = n?'\ n':' '); -         ElsePuts"impossible!"); About     } $     return 0; -}

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Codeforces 321C Ciel the Commander