poj 3107 求树的所有重心

Godfather

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 6123 Accepted: 2165

Description

Last years Chicago was full of gangster fights and strange murders. The chief of the police got really tired of all these crimes, and decided to arrest the mafia leaders.

Unfortunately, the structure of Chicago mafia is rather complicated. There aren persons known to be related to mafia. The police have traced their activity for some time, and know that some of them are communicating with each other. Based on the data collected, the chief of the police suggests that the mafia hierarchy can be represented as a tree. The head of the mafia, Godfather, is the root of the tree, and if some person is represented by a node in the tree, its direct subordinates are represented by the children of that node. For the purpose of conspiracy the gangsters only communicate with their direct subordinates and their direct master.

Unfortunately, though the police know gangsters’ communications, they do not know who is a master in any pair of communicating persons. Thus they only have an undirected tree of communications, and do not know who Godfather is.

Based on the idea that Godfather wants to have the most possible control over mafia, the chief of the police has made a suggestion that Godfather is such a person that after deleting it from the communications tree the size of the largest remaining connected component is as small as possible. Help the police to find all potential Godfathers and they will arrest them.

Input

The first line of the input file contains n — the number of persons suspected to belong to mafia (2 ≤n ≤ 50 000). Let them be numbered from 1 to n.

The following n − 1 lines contain two integer numbers each. The pair a_i, b_i means that the gangstera_i has communicated with the gangsterb_i. It is guaranteed that the gangsters’ communications form a tree.

Output

Print the numbers of all persons that are suspected to be Godfather. The numbers must be printed in the increasing order, separated by spaces.

Sample Input

61 22 32 53 43 6

Sample Output

2 3

Source

Northeastern Europe 2005, Northern Subregion

给定一棵树,求树的所有重心,按照编号从小到大的顺序输出.

分析:本题与上题基本上一样,只是求的量不同,既然我们在找树的重心的时候用的树型dp,而且是求的子树中节点数的最大值,然后求所有最

大值的最小值,那么就有可能存在多个重心,我们每更新到一个最小值的时候就记录其它的最小值也为这个最小值的重心,这样下去就会找到所

有的重心.

 

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <iostream>using namespace std;const int N=200005;int da;int head[N];int ans[N];int son[N];int cnt;int num;struct node{    int v,next;} edge[N];void init(){    cnt=0;    memset(head,-1,sizeof(head));    // memset(vis,0,sizeof(vis));}void add(int u,int v){    edge[cnt].v=v;    edge[cnt].next=head[u];    head[u]=cnt++;}void dfs(int u,int fa,int n){    son[u]=0;    int res=0;    for(int i=head[u]; ~i; i=edge[i].next)    {        int v=edge[i].v;        if(v==fa)            continue;        dfs(v,u,n);        son[u]+=son[v]+1;        res=max(res,son[v]+1);    }    res=max(res,n-son[u]-1);    if(res<da )    {        da=res;        num=1;        ans[0]=u;    }    else if(res==da)        ans[num++]=u;}int main(){    int n;    while(cin>>n)    {        init();        for(int i=1; i<=n-1; i++)        {            int u,v;            scanf("%d%d",&u,&v);            add(u,v);            add(v,u);        }        da=999999999;        dfs(1,-1,n);        sort(ans,ans+num);        for(int i=0; i<num; i++)            printf("%d ",ans[i]);        cout<<"\n";    }    return 0;}