Pseudoforest(伪最大生成树）

Pseudoforest

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

Total Submission(s): 389 Accepted Submission(s): 165

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

 

Source

“光庭杯”第五届华中北区程序设计邀请赛 暨 WHU第八届程序设计竞赛

 

Recommend

lcy

/*
初级想方法，最大生成树再加一条最长边
讲解：没看明白题意的傻逼想法，题目说不是最大生成树，森林也可以，但是最多只能有一个环

正解：将所有的边都加到树上，加的时候如果两点在同一个并查集：如果原来有环就不能加
如果不在同一个并查集：如果原来两个都有环不能加
*/
#include<bits/stdc++.h>
using namespace std;
struct node
{
    int u,v,val;
    node()
    {}
    node(int a,int b,int c)
    {
        u=a;
        v=b;
        val=c;
    }
    bool operator < (const node &a) const
    {
        return val>a.val;
    }
};
vector<node>edge;
int bin[10005];
int n,m;
int x,y,val;
int h[10005];//表示当前集合有没有环
long long cur=0;
void init()
{
    for(int i=0;i<=n;i++)
    {
        bin[i]=i;
        h[i]=0;
    }
    edge.clear();
    cur=0;
}
int findx(int x)
{
    int temp=x;
    while(x!=bin[x])
        x=bin[x];
    bin[temp]=x;
    return x;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
    {
        init();
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&x,&y,&val);
            edge.push_back(node(x,y,val));
        }
        sort(edge.begin(),edge.end());
        for(int i=0;i<edge.size();i++)
        {
            int fx=findx(edge[i].u);
            int fy=findx(edge[i].v);
            if(fx==fy)//两个原来就是一个并查集的就可能产生环了
            {
                if(h[fx])//有环了
                    continue;
                cur+=edge[i].val;
                h[fx]=1;
            }
            else
            {
                if(h[fx]&&h[fy])//两个集合都有环不可以
                    continue;
                bin[fy]=fx;
                cur+=edge[i].val;
                if(h[fx]||h[fy])
                    h[fx]=1;
            }
        }
        printf("%lld\n",cur);
    }
    return 0;
}