F.Three pahs on a tree

 思路

　　两次bfs找出树的直径并处理出端点离树上各叶子节点的距离，在直径上找一点的子树叶子p3，使得dis(p1,p2) + dis(p2,p3) + dis(p1,p3)最大

　　易知上式是路径实长的两倍

#include <bits/stdc++.h>
#define dbg(x) cout << #x << "=" << x << endl
#define eps 1e-8
#define pi acos(-1.0)

using namespace std;
typedef long long LL;

template<class T>inline void read(T &res)
{
    char c;T flag=1;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
    while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
}

namespace _buff {
    const size_t BUFF = 1 << 19;
    char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
    char getc() {
        if (ib == ie) {
            ib = ibuf;
            ie = ibuf + fread(ibuf, 1, BUFF, stdin);
        }
        return ib == ie ? -1 : *ib++;
    }
}

int qread() {
    using namespace _buff;
    int ret = 0;
    bool pos = true;
    char c = getc();
    for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
        assert(~c);
    }
    if (c == '-') {
        pos = false;
        c = getc();
    }
    for (; c >= '0' && c <= '9'; c = getc()) {
        ret = (ret << 3) + (ret << 1) + (c ^ 48);
    }
    return pos ? ret : -ret;
}

const int maxn = 2e5 + 7;

int head[maxn << 1], edge[maxn << 1], nxt[maxn << 1];
int w[maxn << 1];
int vis[maxn];
int dis[maxn];

int n, cnt;

void BuildGraph(int u, int v, int c) {
    cnt++;
    edge[cnt] = v;
    nxt[cnt] = head[u];
    w[cnt] = c;
    head[u] = cnt;
}

int bfs(int x) {
    memset(vis, 0, sizeof(vis));
    memset(dis, 0, sizeof(dis));
    int pos = 0;
    queue <int> q;
    q.push(x);
    vis[x] = 1;
    int u;
    while(!q.empty()) {
        u = q.front();
        //dbg(u);
        q.pop();
        for (int i = head[u]; i; i = nxt[i]) {
            int v = edge[i];
            int d = w[i];
            if(vis[v])
                continue;
            else {
                dis[v] = dis[u] + d;
                vis[u] = 1;
                if(dis[v] > dis[pos]) {
                    pos = v;
                }
                q.push(v);
            }
        }
    }
    return pos;
}

int d1[maxn], d2[maxn];

int main()
{
    read(n);
    int a, b;
    for ( int i = 1; i < n; ++i ) {
        read(a);
        read(b);
        BuildGraph(a, b, 1);
        BuildGraph(b, a, 1);
    }
    int p1, p2;
    p1 = bfs(1);
    p2 = bfs(p1);
    for ( int i = 1; i <= n; ++i ) {
        d1[i] = dis[i];     
    }   
    int p3 = bfs(p2);
    for ( int i = 1; i <= n; ++i ) {
        d2[i] = dis[i];
    }
    int p4 = 0;
    for ( int i = 1; i <= n; ++i ) {
        if(d1[i] + d2[i] > d1[p4] + d2[p4] && i != p1 && i != p2) {
            p4 = i;
        }
    }
    int ans = d1[p4]+d2[p4]+d1[p2];
    //dbg(ans);
    printf("%d
",ans / 2);
    printf("%d %d %d
",p1,p2,p4);
    return 0;
}