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.

 

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 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

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

ac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()  
{  
    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;  
}