1021. Deepest Root (25) 并查集&&DFS

1021. Deepest Root (25)
时间限制
1500 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

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

提交代码

这道题要注意一点要用邻接表存储图，如果用邻接矩阵会超时（稀疏图）

第一次写有返回值的DFS，还不是很熟练

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=10001;
bool vis[maxn];
int a[maxn];
int n;
int cnt=0;
vector<int> ma[maxn];//邻接表
int d[maxn];//计算以这个点为树根时的最大深度
int dfs(int s)//dfs表示以S为起点所能达到的最大深度
{
    int ans=0;
    if(vis[s])return 0;//这个点已经被访问过，明显以这个点的最大深度是0
    vis[s]=true;
    int m=ma[s].size();
    for(int i=0; i<m; i++)
    {
        if(!vis[ma[s][i]])
        {
            int tmp=dfs(ma[s][i]);//以当前点为起点访问能到达的最大的深度，也就是找出S的邻接点里的，深度，能达到的最大的
            ans=max(ans,tmp);//ans记录下最大的深度即可
        }
    }
    return ans+1;//能到这里说明以S点能到达的深度可以加一，也就是S相邻的顶点里的深度，最大值，加上S点，所以就是ans+1
}

void init(int n)//这是并查集的初始化
{
    for(int i=0; i<=n; i++)
        a[i]=i;
}
int find(int x)//并查集查找X的父亲，带路径压缩
{
    if(a[x]!=x)
        a[x]=find(a[x]);
    return a[x];
}
void unio(int x,int y)//合并X,Y到一起
{
    x=find(x);
    y=find(y);
    if(a[x]==a[y])return ;
    a[x]=y;
}

int main()
{
    int i,j,k,t;
    // freopen("in.txt","r",stdin);
    cin>>n;
    init(n);
    for(i=1; i<n; i++)
    {
        int s,e;
        cin>>s>>e;
        unio(s,e);
        ma[s].push_back(e);
        ma[e].push_back(s);
    }

    int sum=0;//判断连通分量的个数
    for(i=1; i<=n; i++)
    {
        if(a[i]==i)sum++;
    }
    if(sum>1)
    {
        printf("Error: %d components\n",sum);
        return 0;
    }
    else
        for(i=1; i<=n; i++)
            {
                memset(vis,0,sizeof(vis));
                d[i]=dfs(i);
            }
    int maxv=-1;int index=0;
    for(i=1;i<=n;i++)if(d[i]>maxv){maxv=d[i];index=i;}
    for(j=1;j<=n;j++)if(d[j]==d[index])
        printf("%d\n",j);


}