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poj 3107 Godfather (树形dp)

来源：程序员人生发布时间：2015-07-29 07:57:57 阅读次数：3810次

Godfather

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5064		Accepted: 1769

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 are n 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 gangster a_i has communicated with the gangster b_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

6 1 2 2 3 2 5 3 4 3 6

Sample Output

2 3

删点，使剩下的分支中最大的节点数最小。

进行深搜，找到每一个节点的子节点构成子树的节点数目，比较得到答案。

#include<stdio.h> #include<math.h> #include<string.h> #include<stdlib.h> #include<algorithm> #include<queue> #include<vector> using namespace std; #define ll long long #define N 51000 #define mem(a,t) memset(a,t,sizeof(a)) vector<int>g[N]; int ans[N]; struct ege { int v,next; }e[N*2]; int head[N]; int cnt,n,n1,mmin; int num[N]; //记录i子树包括节点数 void add(int u,int v) { e[cnt].v=v; e[cnt].next=head[u]; head[u]=cnt++; } int dfs(int u,int fa) { int i,v,tmp=0,t1; num[u]=1; //求当前树枝U的子节点和本身的个数和 for(i=head[u];i!=⑴;i=e[i].next) { v=e[i].v; if(v!=fa) { num[u]+=dfs(v,u); tmp=max(tmp,num[v]); } } t1=max(tmp,n-num[u]); //去掉点u后，最大树的节点数目 if(t1<mmin) { mmin=t1; n1=0; ans[n1++]=u; } else if(t1==mmin) { ans[n1++]=u; } return num[u]; } int main() { //freopen("in.txt","r",stdin); int i,j,u,v; while(~scanf("%d",&n)) { mem(head,⑴); cnt=0; for(i=1;i<n;i++) { scanf("%d%d",&u,&v); add(u,v); add(v,u); } mem(num,0); //num[i]记录节点i的子树点个数 n1=0; mmin=n; dfs(1,⑴); sort(ans,ans+n1); for(i=0;i<n1;i++) printf("%d%c",ans[i],(i==n1⑴)?' ':' '); } return 0; }

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