HDU 3367 Pseudoforest(Kruskal)
HDU 3367 Pseudoforest(Kruskal)
Pseudoforest
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2382 Accepted Submission(s): 933
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 0Sample Output
3
5求出一个最大的子图（子图的每个连通分量最多有一个环） 用kruskal算法求出最大生成树 不过要判断是否有环 2树合并时 ：若2个子树都有环不能合并 只有一个有环可以合并 但合并后的树有环 若2个子树都没环直接合并
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
#define MAX 10000
int n,m;
struct Node
{
int x;
int y;
int w;
}a[MAX*10+5];
int father[MAX+5];
bool tag[MAX+5];
int cmp(Node a,Node b)
{
return a.w>b.w;
}
int find(int x)
{
if(x!=father[x])
father[x]=find(father[x]);
return father[x];
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0&&m==0)
break;
for(int i=0;i<m;i++)
scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].w);
for(int i=0;i<n;i++)
{
father[i]=i;
tag[i]=false;
}
sort(a,a+m,cmp);
int ans=0;
for(int i=0;i<m;i++)
{
int xx=find(a[i].x);
int yy=find(a[i].y);
if(xx!=yy)
{
if(tag[xx]&&tag[yy])
continue;
ans+=a[i].w;
father[xx]=yy;
if(tag[xx]||tag[yy])
{tag[xx]=true;tag[yy]=true;}
}
else
{
if(tag[xx]||tag[yy])
continue;
ans+=a[i].w;
tag[xx]=true;tag[yy]=true;
}
}
printf("%d\n",ans);
}
return 0;
}