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 x and 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.
题目分析 ： 给你一棵树，A,B,C... 表示等级，现要求两个相同等级的之间必须有一个比它大的字母，输出所有的字母序
思路分析 ： 每次找树的重心，将它标记一个字母即可，因为找重心每次是减少一半的点，因此最多可以标记整个树是 2^25 个点的树
代码示例 ：
const int maxn = 1e5+5;
const int inf = 0x3f3f3f3f;
#define ll long long
int num = 0;
vector<int>ve[maxn];
int balance, root;
bool done[maxn];
int size[maxn], mx[maxn];
char ans[maxn];
void getroot(int x, int fa){
size[x] = 1, mx[x] = 0; //以当前结点为根节点的最大结点个数
for(int i = 0; i < ve[x].size(); i++){
int to = ve[x][i];
if (to == fa || done[to]) continue;
getroot(to, x);
size[x] += size[to];
mx[x] = max(mx[x], size[to]);
}
mx[x] = max(mx[x], num-size[x]);
if (mx[x] < balance) {balance = mx[x], root = x;}
}
void dfs(int x, int k){
done[x] = true;
ans[x] = 'A'+k;
for(int i = 0; i < ve[x].size(); i++){
int to = ve[x][i];
if (done[to]) continue;
balance = inf, num = size[to];
getroot(to, to);
dfs(root, k+1);
}
}
int main() {
int n;
int a, b;
cin >> n;
for(int i = 1; i < n; i++){
scanf("%d%d", &a, &b);
ve[a].push_back(b);
ve[b].push_back(a);
}
memset(done, false, sizeof(done));
balance = inf, num = n;
getroot(1, 1);
//printf("root = %d
", root);
dfs(root, 0);
for(int i = 1; i <= n; i++) printf("%c%c",ans[i], i==n?'
':' ');
return 0;
}