Description Given an undirected weighted graph G, you should find one of spanning trees specified as follows. The graph G is an ordered pair (V, E), where V is a set of vertices {v1, v2, …, vn} and E is a set of undirected edges {e1, e2, …, em}. Each…

Slim Span Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7594   Accepted: 4029 Description Given an undirected weighted graph G, you should find one of spanning trees specified as follows. The graph G is an ordered pair (V, E), where V …

Slim Span Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 8633   Accepted: 4608 Description Given an undirected weighted graph G, you should find one of spanning trees specified as follows. The graph G is an ordered pair (V, E), where V …

Slim Span Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 7462   Accepted: 3959 Description Given an undirected weighted graph G, you should find one of spanning trees specified as follows. The graph G is an ordered pair (V, E), where V …

http://poj.org/problem?id=3522 Slim Span Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 5666   Accepted: 2965 Description Given an undirected weighted graph G, you should find one of spanning trees specified as follows. The graph G is a…

先判断是不是连通图,不是就输出-1. 否则,把边排序,从最小的边开始枚举最小生成树里的最短边,对每个最短边用Kruskal算法找出最大边. 或者也可以不先判断连通图,而是在枚举之后如果ans还是INF,说明就没有,就输出-1. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath>…

求一个生成树,使得最大边权和最小边权之差最小.由于数据太小,暴力枚举下界,求出相应的上界.最后取min即可. #include<cstdio> #include<algorithm> #include<cstring> using namespace std; ],rank[]; ;i<=n;i++) fa[i]=i; memset(rank,,sizeof(rank));} int findroot(int x) { if(fa[x]==x) return x;…

kruskal思想,排序后暴力枚举从任意边开始能够组成的最小生成树 #include <cstdio> #include <algorithm> using namespace std; const int maxn = 101; const int maxe = maxn * maxn / 2; struct edge{ int f,t,c; bool operator <(edge e2)const { return c<e2.c; } }e[maxe]; int…

题意: 给你一个图,让你求这个图中所有生成树中满足题目条件的,这个条件是生成树中最长边与最短边的差值最小. 思路: 根据最小瓶颈生成树的定义:在一个有权值的无向图中,求一个生成树最大边的权值尽量小.首先以K算法做这道题,先给所有边排好序,然后我可以从小到大枚举每一条边作为我所求生成树的最短边(即第一次以最短边求最小生成树,第二次删除第一条边,以第二条边为最短边求最小生成树,第三次删除第一条边和第二条边,以第三边为最短边求最小生成树.)然后在这个过程中更新   MST(maxE- minE)就好了…

题意 求生成树的最长边与最短边的差值的最小值 题解 最小生成树保证每一条边最小,就只要枚举最小边开始,跑最小生成树,最后一个值便是最大值 在枚举最小边同时维护差值最小,不断更新最小值. C++代码 /** /*@author Victor /*language C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<…