1021. Deepest Root (25)【并查集+深搜】——PAT (Advanced Level) Practise

题目信息

1021. Deepest Root (25)

时间限制1500 ms 内存限制65536 kB 代码长度限制16000 B

A graph which is connected and acyclic can be considered a tree. The height 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 (<=10000) 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

解题思路

并查集+深搜

AC代码

#include <cstdio>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
int un[10005];
vector<bool> vis(10005);
vector<vector<int> > mp(10005);
int getfa(int a){
return un[a] = (un[a] == a) ? a : getfa(un[a]);
}
void unin(int a, int b){
un[getfa(a)] = un[getfa(b)];
}

int high(int id){
int h = 0;
vis[id] = true;
for (int i = 0; i < mp[id].size(); ++i){
if (!vis[mp[id][i]]){
vis[mp[id][i]] = true;
h = max(h, high(mp[id][i]));
}
}
return h + 1;
}
int main()
{
int n, a, b;
scanf("%d", &n);
for (int i = 1; i <= n; ++i){
un[i] = i;
}
for (int i = 1; i < n; ++i){
scanf("%d%d", &a, &b);
mp[a].push_back(b);
mp[b].push_back(a);
unin(a, b);
}

set<int> st;
for (int i = 1; i <= n; ++i){
st.insert(getfa(i));
}
if (st.size() == 1){
vector<int> v;
int mn = -1;
for (int i = 1; i <= n; ++i){
vis.assign(n+1, false);
int h = high(i);
if (h > mn){
v.clear();
mn = h;
v.push_back(i);
}else if (h == mn){
v.push_back(i);
}
}
for (int i = 0; i < v.size(); ++i){
printf("%d\n", v[i]);
}
}else{
printf("Error: %d components\n", st.size());
}
return 0;
}