「POJ3107」Godfather
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
|
1 2 3 4 5 6 |
6 1 2 2 3 2 5 3 4 3 6 |
Sample Output
|
1 |
2 3 |
题解
这题要求排序输出所有可能的重心
数据卡STL（我也不知道怎么卡的）
所以要边表来存图，用栈来存答案
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 |
#include<map> #include<set> #include<cmath> #include<stack> #include<queue> #include<cstdio> #include<vector> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define mod 998244353 #define pi acos(-1) #define inf 0x7fffffff #define ll long long using namespace std; ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int T,n,mn,top,cnt; int size[50005],mx[50005],ans[50005],last[50005]; struct edge{ int to,next; }e[100005]; void insert(int u,int v) { e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt; e[++cnt].to=u;e[cnt].next=last[v];last[v]=cnt; } void dp(int x,int fa) { size[x]=1; for(int i=last[x];i;i=e[i].next) { int y=e[i].to; if(y==fa)continue; dp(y,x); size[x]+=size[y]; mx[x]=max(mx[x],size[y]); } mx[x]=max(mx[x],n-size[x]); if(mx[x]<mn) { mn=mx[x]; top=0; } if(mx[x]==mn)ans[++top]=x; } int main() { n=read(); mn=inf; for(int i=1;i<n;i++) { int u=read(),v=read(); insert(u,v); } dp(1,0); sort(ans+1,ans+top+1); for(int i=1;i<=top;i++) printf("%d ",ans[i]); return 0; } |
OI 课件=>
博客模板非常不错。