Problem H

最新推荐文章于 2024-05-18 20:34:42 发布

原创最新推荐文章于 2024-05-18 20:34:42 发布 · 315 阅读

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本文介绍了一种使用并查集解决图论问题的方法，旨在找到一个无向图的最大伪森林，即每个连通分量最多只有一个环的子图。通过逐步解析算法流程，包括输入输出样例与AC代码，帮助读者理解如何高效地解决该问题。

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

简单题意：

求出一个图的最大连通子图，连通分量最多只有一个环。现在需要编写一个程序，求出边的权值最大。在输入中给出顶点的个数、边的个数。

解题思路形成过程：

很典型的采用并查集处理回路的问题。和之前的题目很类似，但是还要判断是否有环，并且考虑三种合并并查集的情况。

感想：

题目做的越多，经验就会愈加丰富。

AC代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int visit[10005];
int set[10005];
struct node
{
 int s;
 int e;
 int len;
}a[100005];
int cmp(node a,node b)
{
 return a.len>b.len;
}
int find(int x)
{
 int r=x;
 while(r!=set[r])
 r=set[r];
 int i=x;
 while(i!=r)
 {
 int j=set[i];
 set[i]=r;
 i=j;
 }
 return r;
}
int main()
{
 int n,m;
 while(scanf("%d%d",&n,&m)!=EOF)
 {
 memset(a,0,sizeof(a));
 int ans=0;
 if(n==0&&m==0)
 break;
 for(int i=0;i<=n;i++)
 {
 set[i]=i;
 visit[i]=0;
 }
 for(int i=1;i<=m;i++)
 scanf("%d%d%d",&a[i].s,&a[i].e,&a[i].len);
 sort(a+1,a+m+1,cmp);
 for(int i=1;i<=m;i++)
 {
 int fx=find(a[i].s);
 int fy=find(a[i].e);
 if(fx==fy)
 {
 if(visit[fx]==1)
 continue;
 visit[fx]=1;
 }
 else
 {
 if(visit[fx]==1&&visit[fy]==1)
 continue;
 else if(visit[fx]==0)
 set[fx]=fy;
 else
 set[fy]=fx;
 }
 ans+=a[i].len;
 }
 printf("%d\n",ans);
 }
 return 0;
}