Problem H pseudoforest（伪森林，“最大生成树”）

最新推荐文章于 2020-11-18 21:46:39 发布

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最新推荐文章于 2020-11-18 21:46:39 发布

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分类专栏：数据结构

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本文介绍了一种寻找图中最大伪森林（每连通分量最多含一个环）的算法实现，通过实例输入输出展示了算法流程，并提供了完整的AC代码。

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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

Sample Output

3
5

============================

题意：给出一张图，求该图的最大生成树，且规定每棵树中可以并只能存在一个环。

思路：通过父节点建立一个bool数组标记该联通块是否已经有环。

AC代码：

#include<bits/stdc++.h>
using namespace std;
struct tree
{
    int l,r,v;
}a[100010];
int fat[10001];
int flag[100010];
int father(int x)
{
    if(x!=fat[x]) fat[x]=father(fat[x]);
    return fat[x];
}
void unionn(int x,int y)
{
    fat[x]=y;
}
bool cmp(tree x,tree y)
{
    return x.v>y.v;
}
int main()
{
    int n,m;
    while(cin>>n>>m&&n+m)
    {
        memset(flag,0,sizeof(flag));
        for(int i=0;i<=n;++i)fat[i]=i;
        for(int i=1;i<=m;++i)
        {
            scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].v);
        }
        sort(a+1,a+1+m,cmp);
        int tot=0;
        for(int i=1;i<=m;++i)
        {

            int fa=father(a[i].l),fb=father(a[i].r);

            if(fa!=fb)
            {
               if(flag[fa]&&flag[fb])continue;//若两个联通块均有环，则不能将它们合并；
               tot+=a[i].v;
               //cout<<a[i].l<<" "<<a[i].r<<" "<<i<<"yi"<<endl;
               //cout<<fa<<" "<<fb<<"yi"<<endl;
               unionn(fa,fb);
               if(flag[fa]||flag[fb])//有一者已经成环，另一端也被标记；
               {
                   flag[fa]=1;
                   flag[fb]=1;
               }
            }
            else//当前两点在同一棵树中；
            {
                if(!flag[fa])//若未成环，如下操作；
                {
                    tot+=a[i].v;
                    flag[fa]=1;
                }
            }
        }
        printf("%d\n",tot);
    }
    return 0;
}

The end；