HDU 3367 Pseudoforest 并查集求最大生成树及判断环

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最新推荐文章于 2022-04-05 21:37:50 发布

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本文介绍了一种基于Kruskal算法求解最大伪森林的问题，通过适当修改Kruskal算法来确保每个连通分量最多包含一个环。文章提供了一个具体的实现示例，包括输入输出样例。

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Pseudoforest

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1365 Accepted Submission(s): 514

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

Source

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题意是说给你一个图，可能是非联通的，让你求其一个子图，这个子图的要求是最大生成树，而且最多有一个环。

此题只需在Kruskal的方法上稍加修改即可，因为每个连通分量最多有一个环，所以在合并两棵树的时候要判断是否都存在环，如果都存在，则不可以合并，如果有一棵树有环，则可以合并，并做好标记，如果都没有环，则直接合并。如果是在同一棵树，则要判断这棵树是否存在环，如果不存在则可以合并，并标记好。否则不可以。

PS：万能的主，明天英语四级了，今晚还这a题，保我过吧。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#define M 10007
using namespace std;
int circle[M],pre[M];
int n,m;
struct E
{
    int u,v,value;
}edg[M*20];
int cmp(E a,E b)
{
    return a.value>b.value;
}
void init()
{
    for(int i=0;i<=n;i++)
    {
        pre[i]=i;
        circle[i]=0;
    }
}
int find(int x)
{
    while(x!=pre[x])
        x=pre[x];
    return x;
}
void unio(int a,int b)
{
    int x=find(a);
    int y=find(b);
    if(a==b)return;
    pre[a]=b;
}
int main()
{
    while(scanf("%d%d",&n,&m),n|m)
    {
        init();
        for(int i=0;i<m;i++)
            scanf("%d%d%d",&edg[i].u,&edg[i].v,&edg[i].value);
        sort(edg,edg+m,cmp);
        int count=0;
        for(int i=0;i<m;i++)
        {
            int root_x=find(edg[i].u);
            int root_y=find(edg[i].v);
            if(root_x!=root_y)//如果不是在同一棵树上
            {
                if(circle[root_x]&&circle[root_y])continue;//如果都存在环
                if(circle[root_x]||circle[root_y])circle[root_x]=circle[root_y]=1;//如果有一个存在环
                count+=edg[i].value;//如果都不存在环或者有一个存在环
                unio(root_x,root_y);
            }
            else if(!circle[root_x])//如果是在同一棵树上且此树不存在环
            {
                unio(root_x,root_y);
                circle[root_x]=1;
                count+=edg[i].value;
            }
        }
        printf("%d\n",count);
    }
    return 0;
}