134. Centroid
time limit per test: 0.25 sec. memory limit per test: 4096 KB
You are given an undirected connected graph, with N vertices and N-1 edges (a tree). You must find the centroid(s) of the tree. In order to define the centroid, some integer value will be assosciated to every vertex. Let's consider the vertex k. If we remove the vertex k from the tree (along with its adjacent edges), the remaining graph will have only N-1 vertices and may be composed of more than one connected components. Each of these components is (obviously) a tree. The value associated to vertex k is the largest number of vertices contained by some connected component in the remaining graph, after the removal of vertex k. All the vertices for which the associated value is minimum are considered centroids.
Input
The first line of the input contains the integer number N (1<=N<=16 000). The next N-1 lines will contain two integers, a and b, separated by blanks, meaning that there exists an edge between vertex a and vertex b.
Output
You should print two lines. The first line should contain the minimum value associated to the centroid(s) and the number of centroids. The second line should contain the list of vertices which are centroids, sorted in ascending order.
Sample Input
7 1 2 2 3 2 4 1 5 5 6 6 7Sample Output
3 1 1找树的重心
#include <bits/stdc++.h> using namespace std; const int M=16000+5; int n; int tot=2; int head[M]; int dp[M]; int m[M]; bool vis[M]; int ans=M; int num=1; struct node{ int to; int next; }edge[2*M]; void add_edge(int a,int b){ edge[tot].to=b; edge[tot].next=head[a]; head[a]=tot++; edge[tot].to=a; edge[tot].next=head[b]; head[b]=tot++; } int dfs(int x){ vis[x]=1; int now=1; for(int i=head[x];i!=0;i=edge[i].next){ if(!vis[edge[i].to]){ int mmm=dfs(edge[i].to); dp[x]=max(dp[x],mmm); now+=mmm; } } if(x!=1){ dp[x]=max(dp[x],n-now); } if(ans==dp[x]){ num++; } else{ if(dp[x]<ans){ ans=dp[x]; num=1; } } return now; } int main() { scanf("%d",&n); for(int i=1;i<n;i++){ int a,b; scanf("%d%d",&a,&b); add_edge(a,b); } dfs(1); printf("%d %d\n",ans,num); for(int i=1;i<=n;i++){ if(dp[i]==ans){ printf("%d ",i); } } return 0; }