P1099 树网的核

题面:https://www.luogu.org/problem/P1099

本题直接考虑直径上的点和直径外的点对选定路径的贡献即可.
Code:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<ctime>
using namespace std;
const int N=100005;
int n,s,head[N],dep[N],fa[N],cnt,ans=0x3f3f3f3f;
bool vis[N];
struct Node{
    int u,v,w,nxt;
}edge[N*2];
void add(int u,int v,int w){
    ++cnt;
    edge[cnt].u=u;
    edge[cnt].v=v;
    edge[cnt].w=w;
    edge[cnt].nxt=head[u];
    head[u]=cnt;
}
int bfs(int s){
    queue<int> q;
    int depth=s;
    q.push(s);
    fa[s]=0;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=head[u];i;i=edge[i].nxt){
            int v=edge[i].v;
            if(v!=fa[u]&&!dep[v]){
                dep[v]=dep[u]+edge[i].w;
                fa[v]=u;
                q.push(v);
                depth=dep[depth]<dep[v]?v:depth;
            }
        }
    }
    return depth;
}
int dfs(int u){
    int s=0;
    for(int i=head[u];i;i=edge[i].nxt){
        int v=edge[i].v;
        if(v!=fa[u]&&!vis[v]){
            s=max(dfs(v)+edge[i].w,s);
        }
    }
    return s;
}
int main(){
    int u,v,w,root;
    scanf("%d%d",&n,&s);
    for(int i=1;i<n;i++){
        scanf("%d%d%d",&u,&v,&w);
        add(u,v,w);
        add(v,u,w);
    }
    root=bfs(1);
    memset(dep,0,sizeof(dep));
    memset(fa,0,sizeof(fa));
    root=bfs(root);
    for(int i=root,j=root;i;i=fa[i]){
        while(dep[j]-dep[i]>s){
            j=fa[j];
        }
        ans=min(ans,max(dep[root]-dep[j],dep[i]));
    }
    for(int i=root;i;i=fa[i]){
        vis[i]=1;
    }
    for(int i=root;i;i=fa[i]){
        ans=max(ans,dfs(i));
    }
    printf("%d\n",ans);
    return 0;
}