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
1 #include <bits/stdc++.h> 2 #define dbg(x) cout << #x << "=" << x << endl 3 #define eps 1e-8 4 #define pi acos(-1.0) 5 6 using namespace std; 7 typedef long long LL; 8 9 template<class T>inline void read(T &res) 10 { 11 char c;T flag=1; 12 while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0'; 13 while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag; 14 } 15 16 namespace _buff { 17 const size_t BUFF = 1 << 19; 18 char ibuf[BUFF], *ib = ibuf, *ie = ibuf; 19 char getc() { 20 if (ib == ie) { 21 ib = ibuf; 22 ie = ibuf + fread(ibuf, 1, BUFF, stdin); 23 } 24 return ib == ie ? -1 : *ib++; 25 } 26 } 27 28 int qread() { 29 using namespace _buff; 30 int ret = 0; 31 bool pos = true; 32 char c = getc(); 33 for (; (c < '0' || c > '9') && c != '-'; c = getc()) { 34 assert(~c); 35 } 36 if (c == '-') { 37 pos = false; 38 c = getc(); 39 } 40 for (; c >= '0' && c <= '9'; c = getc()) { 41 ret = (ret << 3) + (ret << 1) + (c ^ 48); 42 } 43 return pos ? ret : -ret; 44 } 45 46 const int maxn = 2e5 + 7; 47 48 int head[maxn << 1], edge[maxn << 1], nxt[maxn << 1]; 49 int w[maxn << 1]; 50 int vis[maxn]; 51 int dis[maxn]; 52 53 int n, cnt; 54 55 void BuildGraph(int u, int v, int c) { 56 cnt++; 57 edge[cnt] = v; 58 nxt[cnt] = head[u]; 59 w[cnt] = c; 60 head[u] = cnt; 61 } 62 63 int bfs(int x) { 64 memset(vis, 0, sizeof(vis)); 65 memset(dis, 0, sizeof(dis)); 66 int pos = 0; 67 queue <int> q; 68 q.push(x); 69 vis[x] = 1; 70 int u; 71 while(!q.empty()) { 72 u = q.front(); 73 //dbg(u); 74 q.pop(); 75 for (int i = head[u]; i; i = nxt[i]) { 76 int v = edge[i]; 77 int d = w[i]; 78 if(vis[v]) 79 continue; 80 else { 81 dis[v] = dis[u] + d; 82 vis[u] = 1; 83 if(dis[v] > dis[pos]) { 84 pos = v; 85 } 86 q.push(v); 87 } 88 } 89 } 90 return pos; 91 } 92 93 int d1[maxn], d2[maxn]; 94 95 int main() 96 { 97 read(n); 98 int a, b; 99 for ( int i = 1; i < n; ++i ) { 100 read(a); 101 read(b); 102 BuildGraph(a, b, 1); 103 BuildGraph(b, a, 1); 104 } 105 int p1, p2; 106 p1 = bfs(1); 107 p2 = bfs(p1); 108 for ( int i = 1; i <= n; ++i ) { 109 d1[i] = dis[i]; 110 } 111 int p3 = bfs(p2); 112 for ( int i = 1; i <= n; ++i ) { 113 d2[i] = dis[i]; 114 } 115 int p4 = 0; 116 for ( int i = 1; i <= n; ++i ) { 117 if(d1[i] + d2[i] > d1[p4] + d2[p4] && i != p1 && i != p2) { 118 p4 = i; 119 } 120 } 121 int ans = d1[p4]+d2[p4]+d1[p2]; 122 //dbg(ans); 123 printf("%d ",ans / 2); 124 printf("%d %d %d ",p1,p2,p4); 125 return 0; 126 }