Pseudoforest
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1957 Accepted Submission(s): 756
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.
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：
在图论中，如果一个森林中有很多连通分量，并且每个连通分量中至多有一个环，那么这个森林就称为伪森林。
现在给出一个森林，求森林包含的最大的伪森林，
直接按照求最大生成树的思想去搞啊
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;
int n,m,fa[10005];
bool vis[10005];
struct node
{
int x,y,w;
}e[100005];
int find(int x)
{
if(x!=fa[x])
fa[x]=find(fa[x]);
return fa[x];
}
bool cmp(node a,node b)
{
return a.w>b.w;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0&&m==0)
break;
int ans=0;
for(int i=0;i<n;i++)
fa[i]=i,vis[i]=0;
for(int i=1;i<=m;i++)
scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].w);
sort(e+1,e+1+m,cmp);
for(int i=1;i<=m;i++)
{
int fx,fy;
fx=find(e[i].x),fy=find(e[i].y);
if(fx!=fy)
{
if(!vis[fx]&&!vis[fy])
{
ans+=e[i].w;
fa[fx]=fy;
}
else if(!vis[fx]||!vis[fy])
{
ans+=e[i].w;
vis[fx]=vis[fy]=1;
}
}
else
{
if(!vis[fx]&&!vis[fy])
{
ans+=e[i].w;
vis[fx]=vis[fy]=1;
}
}
}
printf("%d
",ans);
}
return 0;
}