CF1777E.Edge Reverse
普及/提高-
通过率:0%
AC君温馨提醒
该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。
题目描述
You will be given a weighted directed graph of n nodes and m directed edges, where the i -th edge has a weight of wi ( 1≤i≤m ).
You need to reverse some edges of this graph so that there is at least one node in the graph from which every other node is reachable. The cost of these reversals is equal to the maximum weight of all reversed edges. If no edge reversal is required, assume the cost to be 0 .
It is guaranteed that no self-loop or duplicate edge exists.
Find the minimum cost required for completing the task. If there is no solution, print a single integer −1 .
输入格式
Each test contains multiple test cases. The first line contains the number of test cases t ( 1≤t≤105 ). The description of the test cases follows.
Each test case begins with a line containing two integers n and m ( 2≤n≤2⋅105 , 1≤m≤2⋅105 ) — the number of nodes in the graph and the number of edges in the graph.
The next m lines of each test case contain 3 integers each — u , v , w ( 1≤u,v≤n , 1≤w≤109 ), indicating an edge from u to v with a weight of w . It is guaranteed that no edge connects a vertex to itself, and no pair of edges share the same origin and destination simultaneously.
It is guaranteed that the sum of n and the sum of m over all test cases do not exceed 2⋅105 .
输出格式
For each test case, output the minimum cost. If there is no solution, print −1 .
输入输出样例
输入#1
4 2 1 1 2 3 5 4 1 2 10 2 3 10 3 1 10 4 5 10 4 5 1 2 10000 2 3 20000 1 3 30000 4 2 500 4 3 20 4 5 1 2 10000 2 3 20000 1 3 30000 4 2 5 4 3 20
输出#1
0 -1 20 5
说明/提示
In the first test case, an edge exists from 1 to 2 , so all nodes are reachable (from 1 ).
In the second test case, no nodes are reachable from any node no matter what edges we reverse, so the answer is −1 .
In the third test case, reversing the 4 -th or 5 -th edge allows all nodes to be reachable from 1 . We choose the 5 -th edge here because its weight is smaller.