1021. Deepest Root (25)

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

#include<iostream>
using namespace std;
struct G{
int node;
G* Next;
} graph[10001];
int count=0;
struct disj{
int no;
disj* father;
}dis[10001];
disj* find(disj* a){
if(a->father!=NULL)
return a->father=find(a->father);
else
return a;
}
void unino(disj* a,disj* b){
if(find(a)==find(b))
count++;
else
a->father=b;
}
void add(int a,int b){
G* t=&graph[a];
G* te;
while(t->Next!=NULL){
t=t->Next;
if(t->node==b)
return;
}
if(a>b)
unino(&dis[a],&dis[b]);
te=new G;
te->Next=NULL;
te->node=b;
t->Next=te;
}
int isread[10001];
int DFS(int a,int N){
isread[a]=1;
G* t=&graph[a];
int max=-1,v;
while(t->Next!=NULL){
t=t->Next;
if(isread[t->node]==0){
v=DFS(t->node,N);
if(v>max)
max=v;
}
}
return max+1;
}
int main(){
int N,max=-1;
cin>>N;int j;
int *h=new int
;
int i,temp1,temp2,temp3;
for(i=0;i<10001;i++){
graph[i].Next=NULL;
graph[i].node=i;
dis[i].father=NULL;
dis[i].no=i;
}
for(i=0;i<N-1;i++){
cin>>temp1>>temp2;
add(temp1,temp2);
add(temp2,temp1);
}
if(count!=0){
cout<<"Error: "<<count+1<<" components";
return 0;
}
for(i=1;i<N+1;i++){
h[i]=-1;
if(graph[i].Next!=NULL&&graph[i].Next->Next==NULL){
for(j=1;j<N+1;j++)
isread[j]=0;
h[i]=DFS(i,N);
}
if(h[i]>max)
max=h[i];
}
for(i=1;i<N+1;i++){
if(h[i]==max)
cout<<i<<endl;
}
}


感想：1.判断联通环的个数，只要判断某项中两点都已经出现过的次数即可，数组即可实现，我只是为了复习一下并查集。
2。算深度建议用DFS，我也想到了给每个边一个值，初始值为0，每次加入一条边就用DFS给能增加深度的点深度+1，直接DFS更加简便，而且其实只要对只有一条边的点进行DFS就可以了。