第一行是两个整数N（3  N  200000）和M，分别表示居住点总数和街道总数。以下M行，每行给出一条街道的信息。第i+1行包含整数Ui、Vi、Ti（1Ui, Vi  N，1  Ti  1000000000），表示街道i连接居住点Ui和Vi，并且经过街道i需花费Ti分钟。街道信息不会重复给出。

仅包含整数T，即最坏情况下Chris的父母需要花费T分钟才能找到Chris。

4 3
1 2 1
2 3 1
3 4 1

4

#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<queue>
#define int long long
using namespace std;
const int INF = 1e9 + 10, MAXN = 1e6 + 10;
inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;
}
int N, M;
struct Edge {int u, v, w, nxt;
}E[MAXN];
int head[MAXN], num = 1;
inline void AddEdge(int x, int y, int z) {E[num] = (Edge) {x, y, z, head[x]};head[x] = num++;
}
int dis[MAXN], mx, mxdis;
void dfs(int x, int fa) {for(int i = head[x]; i != -1; i = E[i].nxt) {if(E[i].v == fa) continue;dis[E[i].v] = dis[x] + E[i].w;dfs(E[i].v, x);if(dis[E[i].v] > mxdis) mxdis = dis[E[i].v], mx = E[i].v;}
}
int Node1, Node2, dis1[MAXN], dis2[MAXN];
int GetAns() {memset(dis, 0, sizeof(dis)); dfs(Node1, 0);for(int i = 1; i <= N; i++) dis1[i] = dis[i];memset(dis, 0, sizeof(dis)); dfs(Node2, 0);for(int i = 1; i <= N; i++) dis2[i] = dis[i];int rt = 0;for(int i = 1; i <= N; i++)rt = max(rt, min(dis1[i], dis2[i]));return rt;
}
main() {
#ifdef WIN32freopen("a.in", "r", stdin);
#endifmemset(head, -1, sizeof(head));N = read(); M = read();for(int i = 1; i <= M; i++) {int x = read(), y = read(), z = read();AddEdge(x, y, z);AddEdge(y, x, z);}dfs(1, 0); Node1 = mx;memset(dis, 0, sizeof(dis)); mxdis = 0;dfs(mx, 0); Node2 = mx;int ans = dis[mx];printf("%I64d", ans + GetAns());
}