HDU - 2196 Computer

Posted 2021-03-07 zzctommy

HDU - 2196 Computer

经发现 ，答案就是到直径2个端点距离的较大值。然后先搞出直径，再从直径2端分别开始，求出每个点与它的距离，取较大值。

#include<bits/stdc++.h>
using namespace std;
const int N=10006;
int n,f[N],ans[N],num_edge,head[N],len,s,t;
struct edge{
    int to,nex,val;
}e[N<<1];
void add(int from,int to,int len)
{
    ++num_edge;
    e[num_edge].nex=head[from];
    e[num_edge].to=to;
    e[num_edge].val=len;
    head[from]=num_edge;
}
void dfs(int u,int fa,int &pos)
{
    if(f[u]>len)len=f[u],pos=u;
    for(int i=head[u];i;i=e[i].nex)
    {
        int v=e[i].to,w=e[i].val;
        if(v==fa)continue;
        f[v]=f[u]+w;
        dfs(v,u,pos);
    }
}
int main()
{
    while(scanf("%d",&n)==1)
    {
        memset(ans,0,sizeof(ans));
        num_edge=0;
        memset(head,0,sizeof(head));
        memset(e,0,sizeof(e));
        for(int i=2,w,v;i<=n;++i)
        {
            scanf("%d%d",&v,&w);
            add(v,i,w);
            add(i,v,w);
        }
        memset(f,0,sizeof(f));len=0;
        dfs(1,-1,s);
        memset(f,0,sizeof(f));len=0;
        dfs(s,-1,t);
        for(int i=1;i<=n;++i)ans[i]=f[i];
        memset(f,0,sizeof(f));len=0;
        dfs(t,-1,s);
        for(int i=1;i<=n;++i)ans[i]=max(ans[i],f[i]);
        for(int i=1;i<=n;++i)printf("%d
",ans[i]);
    }
}