一开始觉得这是个什么神题
然后搜题解发现写什么Floyd+模拟O(n3)。。好**的方法
不想写。。然后写了个O(n)
可过BZOJ加强的500000数据。。
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// <core.cpp> - 03/01/16 20:38:46 // This file is created by XuYike's black technology automatically. // Copyright (C) 2015 ChangJun High School, Inc. // I don't know what this program is. #include <iostream> #include <vector> #include <algorithm> #include <cstring> #include <cstdio> #include <cmath> using namespace std; typedef long long lol; int getint(){ int res=0,fh=1;char ch=getchar(); while((ch>'9'||ch<'0')&&ch!='-')ch=getchar(); if(ch=='-')fh=-1,ch=getchar(); while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar(); return fh*res; } const int MAXN=500001; const int INF=1000000000; vector <pair<int,int> >b[MAXN]; int dis[MAXN],f[MAXN]; bool zj[MAXN]; void dfs(int x){ for(int i=0;i<b[x].size();i++){ int next=b[x][i].first; if(next==f[x]||zj[next])continue; f[next]=x; dis[next]=dis[x]+b[x][i].second; dfs(next); } } int main(){ freopen("core.in","r",stdin); freopen("core.out","w",stdout); int n=getint(),s=getint(); for(int i=1;i<=n-1;i++){ int u=getint(),v=getint(),w=getint(); b[u].push_back(make_pair(v,w)); b[v].push_back(make_pair(u,w)); } int l=1,l2=1; dis[1]=0;dfs(1); for(int i=1;i<=n;i++){f[i]=0;if(dis[i]>dis[l])l=i;}//得到一个端点 dis[l]=0;dfs(l); for(int i=1;i<=n;i++)if(dis[i]>dis[l2])l2=i;//得到第二个端点 int ans=INF,j=l2; for(int i=l2;i;i=f[i]){ while(f[j]&&dis[i]-dis[f[j]]<=s)j=f[j]; ans=min(ans,max(dis[j],dis[l2]-dis[i]));//自行脑补 } for(int i=l2;i;i=f[i])zj[i]=1; for(int i=l2;i;i=f[i])dis[i]=0,dfs(i); for(int i=1;i<=n;i++)ans=max(ans,dis[i]);//自行脑补 printf("%d",ans); return 0; } |
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