BZOJ3124 SDOI2013直径

　　本以为必有高论，结果是个思博题。随便找一条直径，最后答案肯定是这条直径上的连续一段，如果某分支长度等于直径上某端的长度这一端都要被剪掉。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
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 read()
{
    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],fa[N],son[N],t,cnt,root,x;
ll deep[N],ans;
struct data{int to,nxt,len;
}edge[N<<1];
void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}
void dfs(int k,int from)
{
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from)
    {
        deep[edge[i].to]=deep[k]+edge[i].len;
        fa[edge[i].to]=k;
        dfs(edge[i].to,k);
    }
}
ll findmax(int k)
{
    ll mx=deep[k];
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k]) mx=max(mx,findmax(edge[i].to));
    return mx;
}
void solve(int k)
{
    ll mx=0;
    for (int i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=fa[k]&&edge[i].to!=son[k]) mx=max(mx,findmax(edge[i].to));
    if ((deep[k]<<1)==mx&&mx) cnt=0;
    if (mx==ans) {cout<<cnt<<endl;exit(0);}
    if (son[k]) cnt++,solve(son[k]);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj3124.in","r",stdin);
    freopen("bzoj3124.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    n=read();
    for (int i=1;i<n;i++)
    {
        int x=read(),y=read(),z=read();
        addedge(x,y,z),addedge(y,x,z);
    }
    deep[1]=0;dfs(1,1);
    root=1;for (int i=2;i<=n;i++) if (deep[i]>deep[root]) root=i;
    deep[root]=0;fa[root]=0;dfs(root,root);
    x=1;for (int i=2;i<=n;i++) if (deep[i]>deep[x]) x=i;
    ans=deep[x];cout<<ans<<endl;
    while (fa[x]) son[fa[x]]=x,x=fa[x];
    solve(root);
    return 0;
}