给出一棵树 ，和边的权值

求权值最长的一条直径

两次bfs求

第一次以任意点开始 BFS求出第一个端点

第二次以第一次得到的端点 BFS求出第二个端点

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <cctype>
#include <cmath>
#include <string>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
typedef long long LL;
typedef long double LD;
#define pi acos(-1.0)
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
typedef pair<int, int> PI;
typedef pair<int, PI> PP;
#ifdef _WIN32
#define LLD "%I64d"
#else
#define LLD "%lld"
#endif
const int MAXN = 200100;
const int INF = 999999;
//LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
//inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}
//inline void print(LL x){printf(LLD, x);puts("");}
//inline void read(int &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}
struct Edge
{int to,next,val;
} edge[MAXN*2];
int head[MAXN],d[MAXN],tol;
bool vis[MAXN];
void init()
{tol=0;memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)
{edge[tol].to=v,edge[tol].val=w,edge[tol].next=head[u];head[u]=tol++;edge[tol].to=u,edge[tol].val=w,edge[tol].next=head[v];head[v]=tol++;
}
int bfs(int u)
{int point,big=0;memset(vis,false,sizeof(vis));queue<int>q;q.push(u);vis[u]=true;d[u]=0;while(!q.empty()){u=q.front();q.pop();for(int i=head[u]; ~i; i=edge[i].next){int v=edge[i].to;if(!vis[v]){d[v]=d[u]+edge[i].val;if(d[v]>big){big=d[v];point=v;}vis[v]=true;q.push(v);}}}return point;
}
int main()
{
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);
#endifint t,a,b,c,n,m;while(scanf("%d%d",&n,&m)!=EOF){init();for(int i=0; i<m; i++){scanf("%d%d%d %*c",&a,&b,&c);addedge(a,b,c);}int u=bfs(1);int v=bfs(u);int len=d[v];cout<<len<<endl;}return  0;
}