Pseudoforest
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3019 Accepted Submission(s): 1194
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
一开始以为是判断如果两个的父节点相等，并且之前没有出现过这种情况就可以相加,然后之后就再也不用判断两个点的父节点的相等的这个情况,结果wa，以为是练算法所以就看了题解,原来这种投机取钱的方法是错的,如果有两个子集但是他们是两个环就不可以合并，这点需要判断,所以需要详细的判断当前的子集是否有环,这里用chun[]这个数组来判断就好了。
#include <stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int bin[100010],flag;
long long sum;
struct node{
int s,e,w;
}a[100001];
int judge[100001];
int findx(int x)
{
return bin[x]==x?x:bin[x]=findx(bin[x]);
}
int merge(int a,int b,int c)
{
int fx=findx(a);
int fy=findx(b);
if(fx!=fy)
{
if(judge[fx]&&judge[fy])
return 0;
bin[fy]=fx;
sum+=c;
if(judge[fx]||judge[fy])
judge[fx]=1;
}
else if(fx==fy&&!judge[fx])
{
judge[fx]=1;
sum+=c;
}
}
int cmp(node a,node b)
{
return a.w>b.w;
}
int main(int argc, char *argv[])
{
int n,m;
while(scanf("%d %d",&n,&m),m+n)
{
int i;
memset(judge,0,sizeof(judge));
for(i=0;i<n;i++)
{
bin[i]=i;
}
for(i=0;i<m;i++)
{
scanf("%d %d %d",&a[i].s,&a[i].e,&a[i].w);
}
sort(a,a+m,cmp);
sum=0;
for(i=0;i<m;i++)
{
merge(a[i].s,a[i].e,a[i].w);
}
printf("%lld\n",sum);
}
return 0;
}