Luogu4408 [NOI2003]逃学的小孩 （树的直径）

一边一定是直径，另一边从两端点走取最小值的最大值

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long

#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("inn.txt","r",stdin);

#else

#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 200007;

int n;

struct Edge{
	int nxt, pre, w;
}e[N << 1];
int head[N], cntEdge;
inline void add(int u, int v, int w){
	e[++cntEdge] = (Edge){ head[u], v, w}, head[u] = cntEdge;
}

long long dis[N], tmp[N];
int vis[N];
int q[N], top;
inline int BFS(int u){
	R(i,1,n) dis[i] = vis[i] = 0;
    top = 0;
    q[++top] = u;
	vis[u] = 1;
    int maxx = 0, pos;
    while(top){
    	int u = q[top--];
    	for(register int i = head[u]; i; i = e[i].nxt){
    		int v = e[i].pre;
    		if(vis[v]) continue;
    		vis[v] = 1;
    		dis[v] = dis[u] + e[i].w;
    		q[++top] = v;
    		if(dis[v] > maxx){
    			maxx = dis[v];
    			pos = v;
    		}
    	}
    }
    return pos;
}

int main(){
	int m;
	io >> n >> m;
	R(i,1,m){
		int u, v, w;
		io >> u >> v >> w;
		add(u, v, w);
		add(v, u, w);
	}

	int l = BFS(1);
	int r = BFS(l);

	long long ans = dis[r];

	R(i,1,n){
		tmp[i] = dis[i];
	}

	BFS(r);
	long long maxx = 0;
	R(i,1,n){
		maxx = Max(maxx, Min(dis[i], tmp[i]));
	}

	printf("%lld", ans + maxx);

	return 0;
}