hdu 3367 Pseudoforest （最大生成树 最多存在一个环）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3367

Pseudoforest

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2870    Accepted Submission(s): 1126

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

 题目大意：在一个无向图中，给定一些边的联通情况以及边的权值，求最大生成树（最多存在一条环路）。

解题思路：用kruskal的方法按照求最大生成树那样求的，只不过要加一个判断，就是判断两颗子树是够成环，

　　　　　如果各成环，就不能合并，如果只有其中一个成环或者都不成环，那么就可以合并，并对其进行标记。。。

AC代码：

20041234    2017-03-08 16:17:45    Accepted    3367    546MS    2668K    1272 B    G++

#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

struct point 
{
    int u,v,l;
}p[100010];
int parent[10010],n,m,vis[10010];  // vis数组用来标记是否形成环 
bool cmp(point a, point b)
{
    return a.l > b.l;              // 从大到小排列 
}

int find (int x)
{
    int s,tmp;
    for (s = x; parent[s] >= 0; s = parent[s]);
    while (s != x)
    {
        tmp = parent[x];
        parent[x] = s;
        x = tmp;
    }
    return s;
}
void Union(int A, int B)
{
    int a = find(A), b = find(B);
    int tmp = parent[a]+parent[b];
    if (parent[a] < parent[b])
    {
        parent[b] = a;
        parent[a] = tmp;
    }
    else
    {
        parent[a] = b;
        parent[b] = tmp;
    }
}
int kruskal()
{
    int sum = 0,max = 0;
    sort(p,p+m,cmp);
    memset(vis,0,sizeof(vis));
    memset(parent,-1,sizeof(parent));
    for (int i = 0; i < m; i ++)
    {
        int u = find(p[i].u), v = find(p[i].v);
        if (u != v)
        {
            if (vis[u] && vis[v]) continue;  // 如果两棵子树，各自能够形成一个环，则不合并 
            if (vis[u] || vis[v])            // 如果只有其中一个形成环，或者两个都没形成环，合并同时标记 
                vis[u] = vis[v] = 1;
            max += p[i].l;
            Union(u,v);
        }
        else if(!vis[u] || !vis[v])         // 在同一连通分量内且有一个或者两个都没形成环   合并且标记 
        {
            vis[u] = vis[v] = 1;
            max += p[i].l;
            Union(u,v);
        }
    }
    return max;
}
int main ()
{
    while (scanf("%d%d",&n,&m),n+m!=0)
    {
        for (int i = 0; i < m; i ++)
            scanf("%d%d%d",&p[i].u,&p[i].v,&p[i].l);
        printf("%d
",kruskal());
    }
    return 0;
}