题目链接:https://vjudge.net/problem/UVA-10600 题意:叫我们求出最小生成树的边权之和 和次小生成树的边权之和. 思路:我们可以先求出最小生成树,这个不难,如果要求次小生成树,那么我们肯定是要在最小生成树里面去掉一条边(假设是a),这样就变成两颗生成树了,我们就还要找一条边(假设是b)把这两颗树连接.这里我们需要满足的就是b-a最小(b一定大于a),这样找到的生成树就是次小生成树了是吧. 关键就是怎么更快的找到b和a这两条边的权值,使b-a最小.因为现在已经是最…

题意: 现在给一个图,问最小生成树和次小生成树的权值和是多少; 思路: 求最小生成树的两种方法,次小生成树是交换最小生成树的其中一条边得到的,现在得到了最小生成树,枚举不在次小生成树中的边,再求一边最小生成树,这些最小生成树的最小权值就是次小生成树的权值了; AC代码: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmat…

裸题,上模板就行,注意j  !  =  k #include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<cassert> #include<iomanip> #include<cstdlib> #include&…

ACM Contest and Blackout 题目链接:https://vjudge.net/problem/UVA-10600 Description: In order to prepare the “The First National ACM School Contest” (in 20??) the major of the city decided to provide all the schools with a reliable source of power. (The m…

题目链接:https://vjudge.net/problem/UVA-10600 In order to prepare the “The First National ACM School Contest” (in 20??) the major of the city decided to provide all the schools with a reliable source of power. (The major is really afraid of blackoutsJ).…

题目大意:给一张无向图,找出最小生成树和次小生成树. 题目分析:模板题...方法就是枚举所有的比最小生成树中两端点之间的最长边还要长的边,用它替换,再取一个最小的值便是次小生成树了. 代码如下: # include<iostream> # include<cstdio> # include<cstring> # include<cstring> using namespace std; # define REP(i,s,n) for(int i=s;i<…

题目大意: 有n个城市,秦始皇要修用n-1条路把它们连起来,要求从任一点出发,都可以到达其它的任意点.秦始皇希望这所有n-1条路长度之和最短.然后徐福突然有冒出来,说是他有魔法,可以不用人力.财力就变出其中任意一条路出来. 秦始皇希望徐福能把要修的n-1条路中最长的那条变出来,但是徐福希望能把要求的人力数量最多的那条变出来.对于每条路所需要的人力,是指这条路连接的两个城市的人数之和. 最终,秦始皇给出了一个公式,A/B,A是指要徐福用魔法变出的那条路所需人力, B是指除了徐福变出来的那条之外的所…

The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28673   Accepted: 10239 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undir…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 22335   Accepted: 7922 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undirected graph G =…

题意:求一幅无向图的最小生成树与最小生成树,不存在输出-1 解法:用Kruskal求最小生成树,标记用过的边.求次小生成树时,依次枚举用过的边,将其去除后再求最小生成树,得出所有情况下的最小的生成树就是次小的生成树.可以证明:最小生成树与次小生成树之间仅有一条边不同.不过这样复杂度有点高,可达O(m^2). 求次小生成树有O(mlogm+n^2)的算法,详见 代码: #include <iostream> #include <cstdio> #include <cstring…