Building Fire Stations
Time Limit: 5000ms
Memory Limit: 131072KB
This problem will be judged on ZJU. Original ID: 3820
64-bit integer IO format: %lld Java class name: Main
Special Judge
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
2 4 1 2 1 3 1 4 5 1 2 2 3 3 4 4 5
Sample Output
1 1 2 1 2 4
Source
The 2014 ACM-ICPC Asia Mudanjiang Regional Contest
Author
YU, Xiaoyao; ZHUANG, Junyuan
解题：树的直径+二分
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 200010;
4 struct arc{
5 int to,next;
6 arc(int x = 0,int y = -1){
7 to = x;
8 next = y;
9 }
10 }e[maxn<<1];
11 int head[maxn],d[maxn],p[maxn],dT[maxn],dS[maxn],tot,n,S,T;
12 bool vis[maxn];
13 void add(int u,int v){
14 e[tot] = arc(v,head[u]);
15 head[u] = tot++;
16 }
17 queue<int>q;
18 int bfs(int u){
19 while(!q.empty()) q.pop();
20 memset(d,-1,sizeof d);
21 memset(p,-1,sizeof p);
22 d[u] = 0;
23 int ret = u;
24 q.push(u);
25 while(!q.empty()){
26 u = q.front();
27 q.pop();
28 if(d[ret] < d[u]) ret = u;
29 for(int i = head[u]; ~i; i = e[i].next){
30 if(d[e[i].to] == -1){
31 d[e[i].to] = d[u] + 1;
32 p[e[i].to] = u;
33 q.push(e[i].to);
34 }
35 }
36 }
37 return ret;
38 }
39 int depth[maxn];
40 int bfs2(int u){
41 while(!q.empty()) q.pop();
42 q.push(u);
43 int ret = depth[u] = 0;
44 while(!q.empty()){
45 u = q.front();
46 q.pop();
47 ret = max(ret,depth[u]);
48 for(int i = head[u]; ~i; i = e[i].next){
49 if(vis[e[i].to]) continue;
50 depth[e[i].to] = depth[u] + 1;
51 q.push(e[i].to);
52 vis[e[i].to] = true;
53 }
54 }
55 return ret;
56 }
57 bool check(int x){
58 int u = dT[x],v = dS[x],tmp = d[T] - (x<<1),cnt = 0;
59 while(u != v){
60 if(depth[u] + min(cnt,tmp - cnt) > x) return false;
61 u = p[u];
62 ++cnt;
63 }
64 return true;
65 }
66 int main(){
67 int kase,u,v;
68 scanf("%d",&kase);
69 while(kase--){
70 scanf("%d",&n);
71 memset(head,-1,sizeof head);
72 memset(vis,false,sizeof vis);
73 int cnt = tot = 0;
74 for(int i = 1; i < n; ++i){
75 scanf("%d%d",&u,&v);
76 add(u,v);
77 add(v,u);
78 }
79 u = T = bfs(S = bfs(1));
80 while(~u){
81 dT[cnt] = u;
82 dS[d[T] - cnt++] = u;
83 vis[u] = true;
84 u = p[u];
85 }
86 u = T;
87 while(~u){
88 depth[u] = bfs2(u);
89 u = p[u];
90 }
91 int low = 0,high = d[T]>>1,ret = 0;
92 while(low <= high){
93 int mid = (low + high)>>1;
94 if(check(mid)){
95 ret = mid;
96 u = dS[mid];
97 v = dT[mid];
98 if(u == v) u = p[u];
99 high = mid - 1;
100 }else low = mid + 1;
101 }
102 printf("%d %d %d\n",ret,u,v);
103 }
104 return 0;
105 }