Pseudoforest (并查集处理回路问题)

点击打开链接

http://acm.hdu.edu.cn/showproblem.php?pid=3367

1、题目大意：

 

求一个图的最大联通子图，要求每个联通分量最多只有一个环，且所求的边的权值之和最大，

输入包括多组样例，每个样例第一行包含n/m两个整数，分别代表图中顶点的个数，边的个数。接下来的m行，每行有三个整数，分别表示一条边的起点和终点及权值，不存在圈不存在重复边，输入00结束

每组样例输出一个整形数，表示最大权值之和

2、思路：

每输入一条边，判断此边两端点是不是在同一颗树上，如果在同一颗树上，判断树是不是有环，如果有环，则不加入此边，如果没环，加入此边（合并）；

如果两棵树都没有环，直接合并即可，

如果只有一棵树有环，可以合并，并标记，

如果都有环，显然不能合并

3、题目：

 

Pseudoforest 

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 1

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
 

4、代码：

#include
#include
#include
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()
{
    //freopen("E:\acm寒假集训\测试样例\测试数据.txt","r",stdin);
    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++)//初始化错了i=0;
        {
            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)//存在环
            {
                if(visit[fx]==1)
                continue;
                visit[fx]=1;//标记根节点，此树已存在环
            }
            else
            {
                if(visit[fx]==1&&visit[fy]==1)//如果2棵树都存在环，跳出
                continue;
                else if(visit[fx]==0)//如果只有一颗树有环，可以合并
                    set[fx]=fy;
                else//如果2棵树都没环，显然可以合并
                set[fy]=fx;
            }
            ans+=a[i].len;
        }
        printf("%d\n",ans);
    }
    return 0;
}
/*
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
*/