Problem H

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.<br><br>

 

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.<br>The last test case is followed by a line containing two zeros, which means the end of the input.<br>

 

Output

Output the sum of the value of the edges of the maximum pesudoforest.<br>

 

Sample Input

  
  

   
   3 3
0 1 1
1 2 1
2 0 1
4 5
0 1 1
1 2 1
2 3 1
3 0 1
0 2 2
0 0

  
  

 

Sample Output

3
5

简单题意：

  求出一个图的最大连通子图，连通分量最多只有一个环。现在需要编写一个程序，求出边的权值最大。在输入中给出顶点的个数、边的个数。

解题思路形成过程：

  很典型的采用并查集处理回路的问题。和之前的题目很类似，但是还要判断是否有环，并且考虑三种合并并查集的情况。

感想：

  题目做的越多，经验就会愈加丰富。

AC代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int visit[10005];
int set[10005];
struct node
{
    int s;
    int e;
    int len;
}a[100005];
int cmp(node a,node b)
{
    return a.len>b.len;
}
int find(int x)
{
    int r=x;
    while(r!=set[r])
    r=set[r];
    int i=x;
    while(i!=r)
    {
        int j=set[i];
        set[i]=r;
        i=j;
    }
    return r;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(a,0,sizeof(a));
        int ans=0;
        if(n==0&&m==0)
        break;
        for(int i=0;i<=n;i++)
        {
            set[i]=i;
            visit[i]=0;
        }
        for(int i=1;i<=m;i++)
              scanf("%d%d%d",&a[i].s,&a[i].e,&a[i].len);
              sort(a+1,a+m+1,cmp);
        for(int i=1;i<=m;i++)
        {
            int fx=find(a[i].s);
            int fy=find(a[i].e);
            if(fx==fy)
            {
                if(visit[fx]==1)
                continue;
                visit[fx]=1;
            }
            else
            {
                if(visit[fx]==1&&visit[fy]==1)
                continue;
                else if(visit[fx]==0)
                    set[fx]=fy;
                else
                set[fy]=fx;
            }
            ans+=a[i].len;
        }
        printf("%d\n",ans);
    }
    return 0;
}