Paths and Trees time limit per test3 seconds memory limit per test256 megabytes Little girl Susie accidentally found her elder brother's notebook. She has many things to do, more important than solving problems, but she found this problem too interes…

E. Paths and Trees time limit per test: 3 seconds memory limit per test: 256 megabytes input: standard input output: standard output Little girl Susie accidentally found her elder brother's notebook. She has many things to do, more important than sol…

题目链接: 题目 E. Paths and Trees time limit per test 3 seconds memory limit per test 256 megabytes inputstandard input outputstandard output 问题描述 Little girl Susie accidentally found her elder brother's notebook. She has many things to do, more important…

E. Paths and Trees time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Little girl Susie accidentally found her elder brother's notebook. She has many things to do, more important than solving…

E. Paths and Trees time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Little girl Susie accidentally found her elder brother's notebook. She has many things to do, more important than solving…

E. Paths and Trees time limit per test:3 seconds memory limit per test:256 megabytes input:standard input output:standard output Little girl Susie accidentally found her elder brother's notebook. She has many things to do, more important than solving…

[题目链接] https://codeforces.com/contest/545/problem/E [算法] 首先求 u 到所有结点的最短路 记录每个节点最短路径上的最后一条边         答案即为以u为根的一棵最短路径生成树 时间复杂度 : O(NlogN) [代码] #include<bits/stdc++.h> using namespace std; ; const long long INF = 1e60; struct edge { int to , w , nxt; }…

题意 求一个生成树,使得任意点到源点的最短路等于原图中的最短路.再让这个生成树边权和最小. http://codeforces.com/contest/545/problem/E 思路 先Dijkstra一下,再对每个点连的边判断是不是最短路上的边,如果是那再贪心取最小的边即可. 代码 #include<bits/stdc++.h> using namespace std; #define ll long long const ll inf=1e18; const int N=6e5+5; s…

题意与分析 题意是这样的,定义一个从某点出发的所有最短路方案中,选择边权和最小的最短路方案,称为最短生成树. 现在求一棵最短生成树,输出总边权和与选取边的编号. 我们首先要明白这样一个结论:对一个图求Dijkstra后,把所有得到的最短路边全部连起来,生成的图一定是一棵树,是不会有环的.原因自己推一下就可以感受到. 那么这样一来,这个树相当于我们在Dijkstra的时候就已经得到了.记录边是Dijkstra的基本操作,而我们只需要考虑一下当最短路相等时谁更优的情况并更新就可以了. 代码 #inc…

[题目大意] 题目将从某点出发的所有最短路方案中,选择边权和最小的最短路方案,称为最短生成树. 题目要求一颗最短生成树,输出总边权和与选取边的编号.[题意分析] 比如下面的数据: 5 5 1 2 2 2 3 2 3 4 16 1 5 18 4 5 2 1 这个图对于从 1 出发,有两种最短路. 这种最短路方案中 dis[2]=2,dis[3]=4,dis[4]=20,dis[5]=18.边权总和 Sum=44 但如果这样选边,1点到各点的距离依然为最短路,但Sum降为了24. 那么如何选择到最优…