Meaning of the questions:
The number of the tree is selected from a point k is less simple path, a path outside the dots to a maximum minimum distance path.
The maximum value of minimum
answer:
- It can be obtained diametrically disprove this simple path in a certain tree
- The two endpoints of the diameter of the tree can be very simple with both sides determined dfs how to obtain information that this chain it? How can the nature of a back end of this chain ran set a flag! ! !
- Longest chain and can maintain a sub-tree for each node in diameter during backtracking
- This information can be processed with a half easily solved
#include<bits/stdc++.h> using namespace std; const int N=2e5+6; int n,m,dis[N],pos,head[N],k,L,R,maxx,flag,maxdis[N],v[N],deep,id[N]; struct Edge{int to,nex,v;}edge[N<<1]; void add(int a,int b,int c){edge[++pos]=(Edge){b,head[a],c};head[a]=pos;} void dfs(int x,int fa) { for(int i=head[x];i;i=edge[i].nex) { int v=edge[i].to; if(v==fa)continue; dis[v]=dis[x]+edge[i].v; if(dis[v]>maxx)maxx=dis[v],flag=v; dfs(v,x); } } void dfs2(int x,int fa,int d) { maxdis[x]=0; for(int i=head[x];i;i=edge[i].nex) { int v=edge[i].to;if(v==fa)continue; dfs2(v,x,d+1); ::v[x]|=::v[v]; if(!::v[v])maxdis[x]=max(maxdis[x],maxdis[v]+edge[i].v); } if(v[x])id[d]=x,deep=max(deep,d); } bool check(int x) { int pos=1,maxx=0; for(int i=deep;i>=1;i--) { if( dis[id[deep] ]-dis[id[i]]>x ) { pos=i+1;break; } } for(int i=pos;i>=max(pos-k+1,1);i--) maxx=max(maxx,maxdis[id[i]]); return max(maxx,dis[id[max(pos-k+1,1)]]-dis[id[1]])<=x; } int main() { cin>>n>>k;int x,y,z; for(int i=1;i<n;i++)scanf("%d%d%d",&x,&y,&z),add(x,y,z),add(y,x,z); dfs(1,0); L=flag; maxx=0; dis[L]=0; dfs(L,0); R=flag; v[R]=1; dfs2(L,0,1); int l=0,r=1e9,ans=0; while(l<=r) { int mid=l+r>>1; if(check(mid))ans=mid,r=mid-1;else l=mid+1; } cout<<years; return 0 ; }