求树的直径和中心(ZOJ3820)
来源:互联网 发布:淘宝运营新手教程视频 编辑:程序博客网 时间:2024/05/21 13:32
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5374
Marjar University is a beautiful and peaceful place. There are N buildings and N - 1 bidirectional roads in the campus. These buildings are connected by roads in such a way that there is exactly one path between any two buildings. By coincidence, the length of each road is 1 unit.
To ensure the campus security, Edward, the headmaster of Marjar University, plans to setup two fire stations in two different buildings so that firefighters are able to arrive at the scene of the fire as soon as possible whenever fires occur. That means the longest distance between a building and its nearest fire station should be as short as possible.
As a clever and diligent student in Marjar University, you are asked to write a program to complete the plan. Please find out two proper buildings to setup the fire stations.
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
The first line contains an integer N (2 <= N <= 200000).
For the next N - 1 lines, each line contains two integers Xi and Yi. That means there is a road connecting building Xi and building Yi (indexes are 1-based).
Output
For each test case, output three integers. The first one is the minimal longest distance between a building and its nearest fire station. The next two integers are the indexes of the two buildings selected to build the fire stations.
If there are multiple solutions, any one will be acceptable.
Sample Input
241 21 31 451 22 33 44 5
Sample Output
1 1 21 2 4题意:给出n个城市和n-1条道路,要选两个城市作为消防站,问建造在那两个城市可以使距离消防站最远的城市的距离最小?
分析:先求出树的直径,然后在直径中间把边切开,成为两个新的树,分别求出两个新树的重心和直径:
#include"string.h"#include"stdio.h"#include"iostream"#include"algorithm"#include"queue"#include"stack"#include"vector"#include"stdlib.h"#include"map"#include"string"#include"math.h"#define M 400009#define inf 10000000#define INF 0x3f3f3f3fusing namespace std;struct node{ int u,v,next;}edge[M];int t,head[M];void init(){ t=0; memset(head,-1,sizeof(head));}void add(int u,int v){ edge[t].u=u; edge[t].v=v; edge[t].next=head[u]; head[u]=t++;}int dis[M],pre[M],n,use[M],a[M],pref[M],a1[M],a2[M],maxi;int bfs(int u,int fuck){ maxi=0; int i; queue<int>q; for(i=1;i<=n;i++) use[i]=0; dis[u]=0; use[u]=1; q.push(u); int ID=u; while(!q.empty()) { u=q.front(); q.pop(); for(i=head[u];~i;i=edge[i].next) { if(i==(fuck^1)||i==fuck)continue; int v=edge[i].v; if(!use[v]) { use[v]=1; dis[v]=dis[u]+1; pre[v]=u; pref[v]=i; if(maxi<dis[v]) { maxi=dis[v]; ID=v; } q.push(v); } } } return ID;}void bary(){ int i; int u=bfs(1,-2); pre[u]=-1; int v=bfs(u,-2); int cnt=0; a[++cnt]=v; for(i=pre[v];i!=-1;i=pre[i]) a[++cnt]=i; int I=pref[a[cnt/2]]; int u1=bfs(a[cnt/2],I); pre[u1]=-1; int v1=bfs(u1,I); int cnt1=0; a1[++cnt1]=v1; for(i=pre[v1];i!=-1;i=pre[i]) a1[++cnt1]=i; int u2=bfs(a[cnt/2+1],I); pre[u2]=-1; int v2=bfs(u2,I); int cnt2=0; a2[++cnt2]=v2; for(i=pre[v2];i!=-1;i=pre[i]) a2[++cnt2]=i; printf("%d %d %d\n",max(cnt1/2,cnt2/2),a1[(cnt1+1)/2],a2[(1+cnt2)/2]);}int main(){ int T,i; cin>>T; while(T--) { scanf("%d",&n); init(); for(i=1;i<n;i++) { int a,b; scanf("%d%d",&a,&b); add(a,b); add(b,a); } bary(); } return 0;}
- 求树的直径和中心(ZOJ3820)
- 2014牡丹江区域赛B(树的直径)ZOJ3820
- zoj3820(树的直径的应用)
- zoj3820 树直径上二分选点
- 求图的绝对中心 && 最小直径生成树 MDST
- 求图的绝对中心 && 最小直径生成树 MDST
- 求树的直径端点和距离
- 求树的直径
- 求树的直径
- 求树的直径
- 求树的直径
- 求树的直径(poj1383验证)
- 求树的直径(最长路径)
- leetcode 刷题(求树的直径)
- 树的直径与树的中心
- 树的直径与树的中心
- 树的直径与树的中心
- BFS求树的直径
- 输入一个8位二进制数,将其转化为十进制数输出
- (二)用JAVA编写MP3解码器——帧头信息解码
- hdu 4939 三色塔防
- 用gSOAP开发Web Service程序
- 奇异值分解(SVD)
- 求树的直径和中心(ZOJ3820)
- python模块安装
- Java二进制协议之Hessian实例(2)
- Linux find 命令的15种经典用法
- 学生类
- HDFS
- 前端case
- (二)用JAVA编写MP3解码器——帧头信息解码
- android -- 蓝牙 bluetooth 入门