题目地址

基本思路：

  • 题目要求树直径的必经边，那么首先应当获取一条直径.
  • 获取直径后从直径上的两个端点分别遍历一次直径，每次遍历直径时从直径上的每个点分别dfs一次并不经过直径上的点，如果深度可以被替换则说明非必经边.
#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
#define ll long long
using namespace std;
const int MAXN=2e5+10,MAXM=MAXN*2;
struct Edge{
	ll from,to,w,nxt;
}e[MAXM];
int head[MAXN],edgeCnt=1;
void addEdge(ll u,ll v,ll w){
	e[++edgeCnt].from=u;
	e[edgeCnt].to=v;
	e[edgeCnt].w=w;
	e[edgeCnt].nxt=head[u];
	head[u]=edgeCnt;
}
ll dep[MAXN];
bool vis[MAXN];
int n;
int from[MAXN];
bool inDiameter[MAXN];//是否在直径上 
ll st,ed;//直径端点 
void markDiameter(){//标记直径 
	int tmp=ed;
	while(tmp){
		inDiameter[tmp]=1;
		tmp=from[tmp];
	}
}
ll bfs(int s){//求直径 
	memset(dep,0,sizeof(dep));
	memset(vis,0,sizeof(vis));
	memset(from,0,sizeof(from));
	queue<int> q;
	q.push(s);
	vis[s]=1;
	while(!q.empty()){
		int nowU=q.front();
		q.pop();
		for(int i=head[nowU];i;i=e[i].nxt){
			int nowV=e[i].to;
			if(!vis[nowV]){
				dep[nowV]=dep[nowU]+e[i].w;
				vis[nowV]=1;
				from[nowV]=nowU;
				q.push(nowV);
			}
		}
	}
	ll ans=0,ansDep=0;
	for(int i=1;i<=n;i++){
		if(dep[i]>ansDep){
			ans=i,ansDep=dep[i];
		}
	}
	return ans;
}
ll dis_fromST[MAXN];//每个点到直径st端点的距离 
bool vis_getDisFromST[MAXN];
void bfs_getDisFromST(){
	queue<int> q;
	q.push(st);
	vis_getDisFromST[st]=1;
	while(!q.empty()){
		int nowU=q.front();
		q.pop();
		for(int i=head[nowU];i;i=e[i].nxt){
			int nowV=e[i].to;
			if(!vis_getDisFromST[nowV]){
				vis_getDisFromST[nowV]=1;
				dis_fromST[nowV]=dis_fromST[nowU]+e[i].w;
				q.push(nowV);
			}
		}
	}
}
ll dis_fromED[MAXN];//每个点到直径ed端点的距离 
bool vis_getDisFromED[MAXN];
void bfs_getDisFromED(){
	queue<int> q;
	q.push(ed);
	vis_getDisFromED[ed]=1;
	while(!q.empty()){
		int nowU=q.front();
		q.pop();
		for(int i=head[nowU];i;i=e[i].nxt){
			int nowV=e[i].to;
			if(!vis_getDisFromED[nowV]){
				vis_getDisFromED[nowV]=1;
				dis_fromED[nowV]=dis_fromED[nowU]+e[i].w;
				q.push(nowV);
			}
		}
	}
}
ll dep_noDiameter[MAXN];//不经过直径的最大深度 
ll dfs_noDiameter(int x,int fa){
	ll maxDep=dep_noDiameter[x];
	for(int i=head[x];i;i=e[i].nxt){
		int nowV=e[i].to;
		if(nowV==fa||inDiameter[nowV])continue;
		dep_noDiameter[nowV]=dep_noDiameter[x]+e[i].w;
		ll tmp=dfs_noDiameter(nowV,x);
		maxDep=max(maxDep,tmp);
	}
	return maxDep;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n-1;i++){
		ll a,b,c;
		cin>>a>>b>>c;
		addEdge(a,b,c);
		addEdge(b,a,c);
	}
	st=bfs(1);
	ed=bfs(st);//直径
	cout<<dep[ed]<<endl;
	markDiameter();//标记直径上点 
	bfs_getDisFromST();
	bfs_getDisFromED();//获取每个点到两个端点的距离 
	int l=st,r=ed;//必经边端点 
	int tmp=from[ed];
	while(tmp){//r
		if(tmp==st)break;
		dep_noDiameter[tmp]=0;
		ll nowMaxDep=dfs_noDiameter(tmp,0);
		if(nowMaxDep==dis_fromED[tmp]){
			r=tmp;
		}
		tmp=from[tmp];
	}
	bfs(ed);//重新获取from数组
	tmp=from[st];
	while(tmp){//l
		if(tmp==ed||tmp==r)break;
		ll nowMaxDep=dfs_noDiameter(tmp,0);
		if(nowMaxDep==dis_fromST[tmp])l=tmp;
		tmp=from[tmp];
	} 
	int cnt=0;
	tmp=l;
	while(tmp){
		if(tmp==r)break;
		cnt++;
		tmp=from[tmp];
	}
	printf("%d\n",cnt);
	return 0;
}