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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!".
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.
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!".
4
1 2
1 3
1 4
A B B B
10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
D C B A D C B D C D
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.
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 }
CodeForces 321C Ciel the Commander
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原文地址：http://www.cnblogs.com/crackpotisback/p/4950886.html
