[TJOI2017]城市

洛咕

分析:(n=5000,3s?)可以直接(n^2)过.枚举删哪条边,这条边的两端分别是一个联通块,求每个联通块以哪个点为根时可以使得 联通块内到根距离最大的点 距离最小,这个点其实就是连通块的直径的中点,可以用(dfs)(O(n))求出直径后,暴力枚举直径上的点找到中点.

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define ll long long
using namespace std;
inline int read(){
    int x=0,o=1;char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    if(ch=='-')o=-1,ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*o;
}
const int N=5005;
int n,maxx,pos,ans=1<<30,dis[N],pre[N];
struct edge{int x,y,z,num;}a[N];
int tot,head[N],nxt[N<<1],to[N<<1],w[N<<1];
inline void add(int a,int b,int c){
	nxt[++tot]=head[a];head[a]=tot;to[tot]=b;w[tot]=c;
}
inline void dfs(int u,int fa,int rt){
	for(int i=head[u];i;i=nxt[i]){
		int v=to[i];if(v==fa||v==rt)continue;
		dis[v]=dis[u]+w[i];pre[v]=u;
		if(dis[v]>maxx)maxx=dis[v],pos=v;
		dfs(v,u,rt);
	}
}
int main(){
	n=read();
	for(int i=1;i<n;++i){
		a[i].num=i;a[i].x=read();a[i].y=read();a[i].z=read();
		add(a[i].x,a[i].y,a[i].z);add(a[i].y,a[i].x,a[i].z);
	}
	for(int i=1;i<n;++i){
		maxx=0;dis[a[i].x]=0;
		dfs(a[i].x,a[i].y,a[i].y);
		maxx=0;dis[pos]=0;int pos1=pos;
		dfs(pos,0,a[i].y);pre[pos1]=0;
		if(maxx>ans)continue;
		int en2=maxx,pos2=pos;
		int now=0,mid=en2/2;
		while(pos2!=pos1){
			if(abs(mid-now)>abs(mid-now-(dis[pos2]-dis[pre[pos2]])))now+=dis[pos2]-dis[pre[pos2]],pos2=pre[pos2];
			else break;
		}
		int hehe1=pos2,hehe2=max(now,en2-now);		
		maxx=0;dis[a[i].y]=0;
		dfs(a[i].y,a[i].x,a[i].x);
		maxx=0;dis[pos]=0;pos1=pos;
		dfs(pos,0,a[i].x);pre[pos1]=0;
		if(maxx>ans)continue;
		int en4=maxx;pos2=pos;
		now=0;mid=en4/2;
		while(pos2!=pos1){
			if(abs(mid-now)>abs(mid-now-(dis[pos2]-dis[pre[pos2]])))now+=dis[pos2]-dis[pre[pos2]],pos2=pre[pos2];
			else break;
		}
		int hehe3=pos2,hehe4=max(now,en4-now);
		ans=min(ans,max(hehe2+hehe4+a[i].z,max(en2,en4)));
	}
	printf("%d
",ans);
    return 0;
}