1021 Deepest Root(25 分)
A graph which is connected and acyclic can be considered a tree. The hight of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤10
4
) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N−1 lines follow, each describes an edge by given the two adjacent nodes’ numbers.
Output Specification:
For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print Error: K components where K is the number of connected components in the graph.
Sample Input 1:
5
1 2
1 3
1 4
2 5
Sample Output 1:
3
4
5
Sample Input 2:
5
1 3
1 4
2 5
3 4
Sample Output 2:
Error: 2 components
题目大意:给出n个结点和n-1条边,问它们能否形成一棵n个结点的树,如果能,从中选出结点作为树根,使整棵树的高度最大。输出所有满足要求的可以作为树根的结点。
思路:
1 由于连通、边数为n-1的图一定是一棵树,因此需要判断给定数据是否能使图连通。使用并查集判断方法:
每读入一条边的两个端点,判断这两个端点是否属于相同的集合,如果不同,则将它们合并到一个集合中,当处理完所有边后根据最终产生的集合个数是否为1来判断给定的图是否连通。
2 确定图连通后,则确定了树,选择合适根结点使树高最大的做法为:
先任意选择一个结点,从该节点开始遍历整棵树,获取能达到的最深的结点,记为集合A;然后从集合A中任意一个结点出发遍历整棵树,获取能达到的最深顶点,记为结点集合B。集合A与B的并集就是所求结果。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 100010;
vector<int> G[maxn];
bool isRoot[maxn];
int father[maxn];
int findFather(int x) {
int a = x;
while(x != father[x]) {
x = father[x];
}
while(a != father[a]) {
int z = a;
a = father[a];
father[z] = x;
}
return x;
}
void Union(int a, int b) {
int faA = findFather(a);
int faB = findFather(b);
if(faA != faB) {
father[faA] = faB;
}
}
void init(int n) {
for(int i = 1; i <= n; i++) {
father[i] = i;
}
}
int blockCnt(int n) {
int cnt = 0;
for(int i = 1; i <= n; i++) {
isRoot[findFather(i)] = true;
}
for(int i = 1; i <= n; i++) {
cnt += isRoot[i];
}
return cnt;
}
int maxH = 0;
vector<int> temp, ans;
void dfs(int u, int height, int pre) {
if(height > maxH) {
temp.clear();
temp.push_back(u);
maxH = height;
} else if(height == maxH) {
temp.push_back(u);
}
for(int i = 0; i < G[u].size(); i++) {
if(G[u][i] == pre) {
continue;
}
dfs(G[u][i], height + 1, u);
}
}
int main() {
int a, b, n;
scanf("%d", &n);
init(n);
for(int i = 1; i < n; i++) {
scanf("%d%d", &a, &b);
G[a].push_back(b);
G[b].push_back(a);
Union(a, b);
}
int block = blockCnt(n);
if(block != 1) {
printf("Error: %d components\n", block);
} else {
dfs(1, 1, -1);
ans = temp;
dfs(ans[0], 1, -1);
for(int i = 0; i < temp.size(); i++) {
ans.push_back(temp[i]);
}
sort(ans.begin(), ans.end());
printf("%d\n", ans[0]);
for(int i = 1; i < ans.size(); i++) {
if(ans[i] != ans[i - 1]) {
printf("%d\n", ans[i]);
}
}
}
return 0;
}