BZOJ4890 Tjoi2017城市

显然删掉的边肯定是直径上的边。考虑枚举删哪一条。然后考虑怎么连。显然新边应该满足其两端点在各自树中作为根能使树深度最小。只要线性求出这个东西就可以了，这与求树的重心的过程类似。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 5010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,p[N],deep[N],fa[N],f[N],len[N],t,root,ans=N*N;
bool flag[N];
struct data{int to,nxt,len;
}edge[N<<1];
void dfs(int k)
{
for (int i=p[k];i;i=edge[i].nxt)
if (edge[i].to!=fa[k])
{
deep[edge[i].to]=deep[k]+edge[i].len;
fa[edge[i].to]=k;
len[edge[i].to]=edge[i].len;
dfs(edge[i].to);
}
}
int dp(int k,int ban)
{
flag[k]=1;
int mx=0,mx2=0,ans=0;
for (int i=p[k];i;i=edge[i].nxt)
if (!flag[edge[i].to]&&edge[i].to!=ban)
{
ans=max(ans,dp(edge[i].to,ban));
int x=f[edge[i].to]+edge[i].len;
if (x>mx) mx2=mx,mx=x;
else if (x>mx2) mx2=x;
}
f[k]=mx;
return max(ans,mx+mx2);
}
int findroot(int k,int ban,int last)
{
int mx=0,mx2=0,l=0,len=0;
for (int i=p[k];i;i=edge[i].nxt)
if (edge[i].to!=fa[k]&&edge[i].to!=ban)
{
int x=f[edge[i].to]+edge[i].len;
if (x>mx) mx2=mx,mx=x,l=edge[i].to,len=edge[i].len;
else if (x>mx2) mx2=x;
}
if (max(last,mx2)+len<mx) return findroot(l,ban,max(last,mx2)+len);
else return k;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("bzoj4890.in","r",stdin);
freopen("bzoj4890.out","w",stdout);
const char LL[]="%I64d\n";
#else
const char LL[]="%lld\n";
#endif
for (int i=1;i<n;i++)
{
}
dfs(1);
int root=1;for (int i=2;i<=n;i++) if (deep[i]>deep[root]) root=i;
fa[root]=deep[root]=0;dfs(root);
int x=1;for (int i=2;i<=n;i++) if (deep[i]>deep[x]) x=i;
while (x!=root)
{
memset(f,0,sizeof(f));
memset(flag,0,sizeof(flag));
int t=max(dp(root,x),dp(x,fa[x]));
int u=findroot(root,x,0),v=findroot(x,fa[x],0);
memset(f,0,sizeof(f));
memset(flag,0,sizeof(flag));
dp(u,x),dp(v,fa[x]);
ans=min(ans,max(f[u]+f[v]+len[x],t));
x=fa[x];
}
cout<<ans;
return 0;
}

posted @ 2018-11-28 19:17  Gloid  阅读(163)  评论(0编辑  收藏  举报