Students find - (diameter of the tree)

 

 

This question is actually a tree without roots of the tree diameter questions:

As shown, first according to the above practice, the tree is determined by two BFS diameter, i.e. AB.

In fact, it was found in practice, problems require max {AB + BC} (BC <AC), or max {AB + AC} (AC <BC), then the two equations can be combined to find max {AB + min {BC , AC}}.

As long as the diameter of the determined endpoint A, B, C then enumerated point, which is the maximum seek answers.

Code:

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<queue>
 5 using namespace std;
 6 typedef long long ll;
 7 const int maxn=200005; 
 8 struct node{
 9     ll to,next,val;
10 }edge[maxn<<1];
11 ll d[maxn],k[maxn],head[maxn],vis[maxn];
12 int n,m,tot;
13 queue<int> q;
14 void add(int u,int v,int w){
15     edge[tot].to=v;
16     edge[tot].next=head[u];
17     edge[tot].val=w;
18     head[u]=tot++;
19 }
20 int BFS(int s){
21     ll maxd=0,Max;
22     memset(d,0,sizeof(d));
23     memset(vis,0,sizeof(vis));
24     while(!q.empty()) q.pop();
25     q.push(s); vis[s]=1;
26     while(!q.empty()){
27         int u=q.front(),v; q.pop();
28         for(int i=head[u];i!=-1;i=edge[i].next){
29             if(vis[v=edge[i].to]) continue;
30             vis[v]=1;
31             d[v]=d[u]+edge[i].val;
32             q.push(v);
33             if(d[v]>maxd){
34                 maxd=d[v];
35                 Max=v;
36             }
37         }
38     }
39     return Max;
40 }
41 int main(){
42 //    freopen("1.in","r",stdin); 
43     memset(head,-1,sizeof(head));
44     scanf("%d%d",&n,&m);
45     for(int i=1;i<=m;i++){
46         ll x,y,z;
47         scanf("%lld%lld%lld",&x,&y,&z);
48         add(x,y,z); add(y,x,z);
49     }
50     int l=BFS(1);//A 
51     int r=BFS(l);//B
52     ll ans=d[r];//直径长度 
53     for(int i=1;i<=n;i++) k[i]=d[i];//k:AC长度 
54     BFS(r);
55     ll M=0;
56     for(int i=1;i<=n;i++) M=max(M,min(d[i],k[i]));// Enumeration point C d: BC length 
57 is      the printf ( " % LLD \ n- " , ANS + M);
 58      return  0 ;
 59 }