HDU 6041 I Curse Myself（仙人掌+tanjan）
I Curse Myself
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 475 Accepted Submission(s): 76
Problem Description
There is a connected undirected graph with weights on its edges. It is guaranteed that each edge appears in at most one simple cycle.
Assuming that the weight of a weighted spanning tree is the sum of weights on its edges, define V(k) as the weight of the k-th smallest weighted spanning tree of this graph, however, V(k) would be defined as zero if there did not exist k different weighted spanning trees.
Please calculate (∑k=1Kk⋅V(k))mod232.
Assuming that the weight of a weighted spanning tree is the sum of weights on its edges, define V(k) as the weight of the k-th smallest weighted spanning tree of this graph, however, V(k) would be defined as zero if there did not exist k different weighted spanning trees.
Please calculate (∑k=1Kk⋅V(k))mod232.
Input
The input contains multiple test cases.
For each test case, the first line contains two positive integers n,m (2≤n≤1000,n−1≤m≤2n−3), the number of nodes and the number of edges of this graph.
Each of the next m lines contains three positive integers x,y,z (1≤x,y≤n,1≤z≤106), meaning an edge weighted z between node x and node y. There does not exist multi-edge or self-loop in this graph.
The last line contains a positive integer K (1≤K≤105).
For each test case, the first line contains two positive integers n,m (2≤n≤1000,n−1≤m≤2n−3), the number of nodes and the number of edges of this graph.
Each of the next m lines contains three positive integers x,y,z (1≤x,y≤n,1≤z≤106), meaning an edge weighted z between node x and node y. There does not exist multi-edge or self-loop in this graph.
The last line contains a positive integer K (1≤K≤105).
Output
For each test case, output "
Case #x: y" in one line (without quotes), where
x indicates the case number starting from
1 and
y denotes the answer of corresponding case.
Sample Input
4 3 1 2 1 1 3 2 1 4 3 1 3 3 1 2 1 2 3 2 3 1 3 4 6 7 1 2 4 1 3 2 3 5 7 1 5 3 2 4 1 2 6 2 6 4 5 7
Sample Output
Case #1: 6 Case #2: 26 Case #3: 493
Source
2017 Multi-University Training Contest - Team 1
Recommend
liuyiding
题目大意：
给你一棵无向带边权的仙人掌（节点数<=1e3），求前k(1<=k<=1e5)小生成树的权值之和。
解题思路：
由于这个图是仙人掌，那么它的生成树就一定是每个环去掉一条边所构成的，我们可以通过存边的tarjan算法找到仙人掌上的所有环，题目要求k小生成树，我们只要找到去掉的边的k大的组合即可。
那么就可把问题简化为：有一些集合，在每个集合中选一个数，求前k大的组合。
官方题解说这就是一个经典问题了。。。可能是我比较菜吧，通过这道题，我才知道这个问题怎么解决。。。。
对于这些集合两两合并（为了方便一个叫A一个叫B），在合并的过程中，先用A中最大值和B中全部元素相加，放到堆中。每次从堆中拿出元素时，用构成这个元素的A中下一个更小的值（如果存在）与之前的B中元素结合放到堆中，直到堆空或得到k个元素。
这样写的复杂度我不是很会推导，引用官方题解：
AC代码：
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