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 
#include 
#include 
#include 
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;
}