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);}