【C++】1021. Deepest Root (25)

1021. Deepest Root (25)

时间限制
1500 ms

内存限制
32000 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 calledthe 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

1.这道题比较适合用bfs，而不是dfs

好吧，20分，有五分没有拿

#include <iostream>
#include <string>
#include<vector>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;
//bool mark[10001];
vector<vector <int>> a;
int level[10001]={0};
queue <int >q;

int bfs(int tt){
level[tt]=1;
q.push(tt);
while(q.size()!=0){
int tt=q.front();
for(int i=0;i<a[tt].size();i++){
if(level[a[tt][i]]==0){
level[a[tt][i]]=level[tt]+1;
q.push(a[tt][i]);
}
}
q.pop();
}

return 0;
}

int maxl(int n){
int max=-1;
for(int i=0;i<n;i++){
if(level[i]>max){
max=level[i];
}
}
return max;
}

int cmp(int a,int b){
return a<b;
}

int main(){
int N,i,j;
freopen("in.txt","r",stdin);
scanf("%d",&N);
a.resize(N);
for(i=0;i<N-1;i++){
int a1,a2;
scanf("%d%d",&a1,&a2);
a1--;
a2--;
a[a1].push_back(a2);
a[a2].push_back(a1);
}
int tt=0;
bfs(tt);
vector <int> maxit;
int suo=maxl(N);
int cc=0;
for( i=0;i<N;i++){
if(level[i]>0){
cc++;
}
if(level[i]==suo){
maxit.push_back(i);
}
}
if(cc==N){
memset(level,0,sizeof(level));
tt=bfs(maxit[0]);
suo=maxl(N);

for( i=0;i<N;i++){

if(level[i]==suo){
maxit.push_back(i);
}
}

sort(maxit.begin(),maxit.end(),cmp);
for(i=0;i<maxit.size();i++){
cout<<maxit[i]+1<<endl;
}
}else{
int count=1;
for(i=0;i<N;i++){
if(level[i]==0){
bfs(i);
count++;
}
}
printf("Error: %d components",count);

}

return 0;
}