6.1.8 Pseudoforest

Pseudoforest

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

Total Submission(s): 182 Accepted Submission(s): 85

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

Sample Output

3
5

思路：伪森林，详见维基百科，其实就是最短路变成最长路，用kruskal

 1 #include <iostream>
 2 #include <cmath>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <cstring>
 6 #include <string>
 7 #include <cstdlib>
 8 using namespace std;
 9 
10 const int maxn=10010,maxm=100100;
11 int p[maxn],fn,fto,n,m,ans;
12 bool f[maxn];
13 struct qq
14 {
15     int n,to,d;
16     friend bool operator < (qq a,qq b)
17     {
18         return a.d>b.d;
19     }
20 } e[maxm];
21 
22 void close()
23 {
24 exit(0);
25 }
26 
27 int getfather(int k)
28 {
29     if (p[k]==k)
30         return k;
31     p[k]=getfather(p[k]);
32     return p[k];
33 }
34 void work()
35 {
36     memset(f,false,sizeof(f));
37     ans=0;
38     for (int i=1;i<=m;i++)
39     {
40         fn=getfather(e[i].n);
41         fto=getfather(e[i].to);
42     //    printf("n:%d fn:%d to:%d fto:%d ans:%d\n",e[i].n,fn,e[i].to,fto,ans);
43         if (fn==fto)
44         {
45             if (not f[fn]) //祖先是同一个，但没有环
46             {
47                 ans+=e[i].d;
48                 f[fn]=true;
49             }
50         }
51         else//没有环，完全可以合并，但要注意有环的赋值
52         {
53             if (f[fn] && f[fto])
54                 continue;
55             ans+=e[i].d;
56             if (f[fn] || f[fto])
57             {
58                 f[fn]=true;
59                 f[fto]=true;
60             }
61                 p[fn]=fto;
62         }
63     }
64     printf("%d\n",ans);
65 }
66 
67 
68 void init()
69 {
70     while (scanf("%d %d",&n,&m)!=EOF)
71     {
72         if (n==0 && m==0) break;
73         for (int i=0;i<=n;i++)
74             p[i]=i;
75         for (int i=1;i<=m;i++)
76         {
77             scanf("%d %d %d",&e[i].n,&e[i].to,&e[i].d);
78         }
79         sort(e+1,e+m+1);
80         work();
81     }
82 }
83 
84 int main ()
85 {
86     init();
87     close();
88     return 0;
89 }