﻿HDU 3367 Pseudoforest-Ocrosoft

# Pseudoforest

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2679    Accepted Submission(s): 1055

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

```

Solution

```#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
#define INF INT_MAX
#define MAXN 10000+10
using namespace std;
int n,m;
struct Edge
{
int u,v,w;
bool operator<(const Edge e)const
{
return w>e.w;//从大到小
}
} edge[100010];
int pre[MAXN];
bool circle[MAXN];
int find(int x)
{
if(pre[x]!=x)pre[x]=find(pre[x]);
return pre[x];
}
int main()
{
while(cin>>n>>m&&(n||m))
{
for(int i=0; i<n; i++)pre[i]=i,circle[i]=0;
for(int i=0; i<m; i++)
scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
sort(edge,edge+m);
int ans=0;
for(int i=0; i<m; i++)
{
int fu=find(edge[i].u),fv=find(edge[i].v);
if(fu!=fv&&(circle[fu]!=1||circle[fv]!=1))//不是同一个集合，并且不是两个都成环
{
pre[fu]=fv;
ans+=edge[i].w;
if(circle[fu])circle[fv]=circle[fu];
}
else if(!circle[fu])//是同一个集合，连上这条边，就成环了
ans+=edge[i].w,circle[fu]=1;
}
printf("%d\n",ans);
}
return 0;
}
```