(hdu step 6.1.8)Pseudoforest(求有一个环的最大生成树)

题目：

Pseudoforest

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 283 Accepted Submission(s): 122 

Problem Description

In graph theory, a pseudoforest is an undirected graph in which every connected component has at most one cycle. The maximal pseudoforests of G are the pseudoforest subgraphs of G that are not contained within any larger pseudoforest of G. A pesudoforest is larger than another if and only if the total value of the edges is greater than another one’s.

 

Input

The input consists of multiple test cases. The first line of each test case contains two integers, n(0 < n <= 10000), m(0 <= m <= 100000), which are the number of the vertexes and the number of the edges. The next m lines, each line consists of three integers, u, v, c, which means there is an edge with value c (0 < c <= 10000) between u and v. You can assume that there are no loop and no multiple edges.
The last test case is followed by a line containing two zeros, which means the end of the input.

 

Output

Output the sum of the value of the edges of the maximum pesudoforest.

 

Sample Input

3 30 1 11 2 12 0 14 50 1 11 2 12 3 13 0 10 2 20 0

 

Sample Output

35

 

 

Source

“光庭杯”第五届华中北区程序设计邀请赛 暨 WHU第八届程序设计竞赛

 

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题目分析：

              求有一个环的最大生成树。

1、要求最大生成树，这时候我们只需要把从小到大的排序函数改成从大到小排序即可。如：

bool cmp(Edge a,Edge b){
return a.weight > b.weight;
}

2、如何保证有一个环？

1）如果是两棵树在进行合并操作

两棵树都有环，不能进行合并操作。

如果只有一个数有环，可以进行合并操作。合并前先做一下标记。

如果都没有环，可以进行合并操作。

2）如果是是同一颗树在进行合并操作。

那么这棵树合并前必然不能有环。

代码如下：

/* * h.cpp * *  Created on: 2015年3月11日 *      Author: Administrator */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 10001;const int maxm = 100001;int father[maxn];struct Edge{int begin;int end;int weight;}edges[maxn*maxn];int visited[maxn];//用来标记以某一结点为根节点的最大生成树中是否已经有环.如visited[i]=true.表示以i为根节点的最大生成树中已经有环int find(int a){if(a == father[a]){return a;}return father[a] = find(father[a]);}/** * 求有一个环的最大生成树 */int kruscal(int count){int sum = 0;int i;for(i = 0 ; i < maxn ; ++i){father[i] = i;}for(i = 0 ; i < count ; ++i){//注意这里的边的索引从0开始,从1开始这道题莫名其妙的WAint fa = find(edges[i].begin);int fb = find(edges[i].end);if(fa != fb){//如果是两棵树if(visited[fa] == true && visited[fb] == true){//并且他们都已经有环continue;//不能合并,继续拧下一次循环}if(visited[fa] || visited[fb]){//如果其中有一棵树有环了.可以继续执行visited[fa] = visited[fb] = true;//合并以后将这两棵树都标记位已经有环}//合并操作sum += edges[i].weight;father[fa] = fb;}else if(visited[fa] == false){//如果是同一棵树,并且还没有环/** * 则能进行合并操作。为什么呢? * 同一棵树的合并操作,必然会导致又多出一个环. * 所以只有在原来还没有环的秦光霞才能继续拧合并 */sum += edges[i].weight;visited[fa] = true;}}return sum;}bool cmp(Edge a,Edge b){return a.weight > b.weight;}int main(){int n,m;while(scanf("%d%d",&n,&m)!=EOF,n){memset(visited,false,sizeof(visited));int i;for(i = 0 ; i < m ; ++i){scanf("%d%d%d",&edges[i].begin,&edges[i].end,&edges[i].weight);}sort(edges,edges+m,cmp);printf("%d\n",kruscal(m));}return 0;}