Luogu1099 树网的核 （暴力？，floyd？）（还未想正解，暴力就A了）

阅读理解两小时，手敲暴力思考5分钟。然后\(n^3\)就A了

暴力代码

#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("in.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 = 307;

int dis[N][N], D[N];
int vis[N];
int main(){
//FileOpen();
	int n, S;
	io >> n >> S;
	R(i,1,n){
		R(j,1,n){
			if(i != j)
				dis[i][j] = 2147483647;
		}
	}
	R(i,2,n){
		int u, v, w;
		io >> u >> v >> w;
		dis[u][v] = dis[v][u] =  w;
	}

	int DIS = 0;
	R(k,1,n){
		R(i,1,n){
			if(dis[i][k] == 2147483647) continue;
			R(j,1,n){
				if(dis[k][j] == 2147483647) continue;
				dis[i][j] = Min(dis[i][j], dis[i][k] + dis[k][j]);
				DIS = Max(DIS, dis[i][j]);
			}
		}
	}

	R(i,1,n){
		R(j,1,n){
			if(dis[i][j] == DIS){
				R(k,1,n){
					if(dis[i][k] + dis[k][j] == dis[i][j]){
						vis[k] = true;
					}
				}
			}
		}
	}

	int ans = 2147483647;
	R(i,1,n){
		R(j,i,n){
			if(vis[i] && vis[j] && dis[i][j] <= S){
				R(xx,1,n) D[xx] = 2147483647;
				R(k,1,n){
					if(dis[i][k] + dis[k][j] == dis[i][j]){
						R(l,1,n){
							D[l] = Min(D[l], dis[k][l]);
						}
					}
				}

				int maxx = 0;
				R(k,1,n){
					maxx = Max(maxx, D[k]);
				}
				ans = Min(ans, maxx);
			}
		}
	}

	printf("%d", ans);

	return 0;
}