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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.<br><br>

 

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.<br>The last test case is followed by a line containing two zeros, which means the end of the input.<br>

 

Output

Output the sum of the value of the edges of the maximum pesudoforest.<br>

 

Sample Input

3 30 1 11 2 12 0 14 50 1 11 2 12 3 13 0 10 2 20 0

 

Sample Output

35

 题目大意：找到一个这样的图，在这个图中，最多有一个环。使得所有的边的和最大。

思路：并查集

#include<iostream>#include<cmath>#include<algorithm>using namespace std;int circle[10005];int total[10006];int n,m,k;struct node{int a;int b;int dis;}p[4000000];bool cmp(node n1,node n2){  return n1.dis>n2.dis;}int parent[10005];int num[10006];void init(int n){for(int i=1;i<=n;i++){  parent[i]=i;  num[i]=1;  circle[i]=0;  total[i]=0;}}int find(int a){    if(a==parent[a])  return a;parent[a]=find(parent[a]);return parent[a];}int findans()  { int ans=0;for( int i=1;i<=m;i++){int a=p[i].a;int b=p[i].b;int p1=find(a);int p2=find(b);if(p1!=p2){ if(circle[p1]==0&&circle[p2]==0) {   parent[p1]=p2;ans+=p[i].dis; } if(circle[p1]==1&&circle[p2]==0) {   parent[p2]=p1;ans+=p[i].dis; } if(circle[p1]==0&&circle[p2]==1) {   parent[p1]=p2;ans+=p[i].dis; }}if(p1==p2&&circle[p1]==0){  total[p1]+=p[i].dis;  circle[p1]=1;  ans+=p[i].dis;}}return ans;}int main(){  while(cin>>n>>m&&n+m)  { if(n==0&&m==0) break;     init(n);int a,b,c;for( int i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);p[i].a=a+1;p[i].b=b+1;p[i].dis=c;}sort(p+1,p+m+1,cmp);   int ans=findans();cout<<ans<<endl;}  return 0;}