Pseudoforest

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

题意：有一个无向图,求它的伪森,要求每一个联通分量最多有一个环,使得伪森林的权值最大

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxh=10000+10;
const int maxe=100000+10;
typedef struct Edge
{
    int u,v,w;
};
Edge E[maxe];
int fa[maxh],cir[maxh],n,m;
void init(int num)
{
    for(int i=0;i<=num;i++)
    {
        fa[i]=i;
        cir[i]=0;
    }
}
int find(int x)
{
    if(x!=fa[x])
    fa[x]=find(fa[x]);
    return fa[x];
}
bool cmp(Edge e1,Edge e2)
{
    return e1.w>e2.w;//降序
}
bool Union(int x,int y)
{
    int a=find(x);
    int b=find(y);
    if(a==b)//若存在环
    {
        if(!cir[a])若之在集合不存在环
        {
            cir[a]=1;//标记a所在集合有环
            return true;
        }
        return false;
    }
    else
    {
        if(cir[a]==cir[b]&&cir[b]==1)若a,b所在集合都有环则不能合并
        return false;
        if(cir[a]==1)若只有集合有环则可以合并或都无环则可以合并
        fa[b]=a;
        else
        fa[a]=b;
        return true;
    }
}
int main()
{
    int ans;
    while(~scanf("%d%d",&n,&m)&&(n+m))
    {
        init(n);
        ans=0;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&E[i].u,&E[i].v,&E[i].w);
        }
        sort(E,E+m,cmp);
        for(int i=0;i<m;i++)
        {
            if(Union(E[i].u,E[i].v))
            ans+=E[i].w;
        }
        printf("%d\n",ans);
    }
    return 0;
}