Godfather - POJ 3107 树形dp
来源:互联网 发布:海文餐饮软件 编辑:程序博客网 时间:2024/05/10 12:15
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 ai, bi means that the gangster ai has communicated with the gangster bi. 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
题意:找出所有删除该点后其他树的最大节点数最小的点。
思路:简单的dfs,但是这道题用vector会超时,应该用静态链表。
AC代码如下:
#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;struct node{ int y,next;}e[100010];int n,ans[50010],minn,num[50010],h[50010],tot;void ins(int u,int v){ e[++tot].y=v; e[tot].next=h[u]; h[u]=tot;}void dfs(int f,int u){ int i,v,maxn=0; num[u]=1; for(i=h[u];i;i=e[i].next) { v=e[i].y; if(v==f) continue; dfs(u,v); num[u]+=num[v]; maxn=max(maxn,num[v]); } maxn=max(maxn,n-num[u]); if(maxn<minn) { minn=maxn; ans[0]=1; ans[1]=u; } else if(maxn==minn) { ans[0]++; ans[ans[0]]=u; }}int main(){ int i,j,k,u,v; while(~scanf("%d",&n)) { memset(h,0,sizeof(h)); tot=0; for(i=1;i<n;i++) { scanf("%d%d",&u,&v); ins(u,v); ins(v,u); } minn=n; ans[0]=0; dfs(0,1); sort(ans+1,ans+1+ans[0]); printf("%d",ans[1]); for(i=2;i<=ans[0];i++) printf(" %d",ans[i]); printf("\n"); }}
- POJ 3107 Godfather(树形dp)
- Godfather - POJ 3107 树形dp
- POJ 3107 Godfather (树形DP)
- poj 3107 Godfather (树形dp)
- 【POJ】3107 Godfather 树形dp
- POJ 3107 - Godfather 树形DP..vector慎用...
- poj 3107 Godfather(树形dp)
- (简单) 树形dp POJ 3107 Godfather
- POJ 3107 Godfather (水题,树形DP)
- POJ 3107 Godfather(树形DP)
- POJ 3107 Godfather(树形DP)
- poj 3107 Godfather(树形DP)
- poj 3107 Godfather(树形dp)
- poj 3170 Godfather(树形dp)
- poj 3107 Godfather(树形dp,树的重心)
- POJ.3107 Godfather (树形DP 树的重心)
- 【树形dp】PKU-3107-Godfather
- 【树形dp+前向星】Godfather POJ
- zsh+oh_my_sh+autojump
- HDU 5042 GCD pair 预处理+二分 分段
- android使用httppost向c# wcf发送数据总结
- metasploit结构剖析
- 深入浅出JPA--映射持久化对象Entity之@Entity
- Godfather - POJ 3107 树形dp
- C# 正则表达式应用积累
- jse基础-进制转换
- Linux shell学习: &&和||
- POJ3009 Curling 2.0(DFS)
- Java回调函数的理解
- ubuntu python小试牛刀---文件内容类型统计
- 关于reverse的问题
- linkin大话设计模式--门面模式