hdu 3367 Pseudoforest

Pseudoforest

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2907    Accepted Submission(s): 1144

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

 

Sample Output

35

 

Source

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题意：在图论中，如果一个森林中有很多连通分量，并且每个连通分量中至多有一个环，那么这个森林就称为伪森林。现在给出一个森林，求森林包含的最大的伪森林，其大小通过所有边的权值之和来比较。

根据题目的要求我们再合并两棵树的时候要判断是否有环：
1.两棵树都有环，那么不能合并（要求最多一个环）
2.如果一棵树有，另一棵没有，那么可以直接合并，并把有环的树标记。
3.如果两棵树都没有环，那么可以直接合并。
如果要查找的这两个点在同一棵树上，并且该数还没有环，那么就把这两点连在一起构成环，标记即可。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#define INF 0x3f3f3f3f#define MAX 1005using namespace std;struct Edge{    int u, v, w;}edge[100004];int vis[10004], pre[10004];int cmp(Edge x, Edge y){    return x.w > y.w;}int Find(int x){    while(x!=pre[x])        x = pre[x];    return x;}int main(){    int n, m;    while(~scanf("%d%d",&n, &m))    {        if(n==0&&m==0)            break;        memset(vis, 0, sizeof(vis));        for(int i = 0; i < n; i++)            pre[i] = i;        for(int i = 0; i < m; i++)        {            scanf("%d%d%d",&edge[i].u, &edge[i].v, &edge[i].w);        }        sort(edge, edge + m, cmp); ///把边排序        int sum = 0;        for(int i = 0; i < m; i++)        {            int a = Find(edge[i].u);            int b = Find(edge[i].v);            if(a!=b) ///a和b不在同一个环中            {                if(!vis[a]&&!vis[b])  ///两棵树都没有环                {                    sum += edge[i].w;                    pre[a] = b;                }                else if(!vis[a])  ///a中没有环                {                    sum += edge[i].w;                    pre[a] = b;                }                else if(!vis[b])///b中没有环                {                    sum += edge[i].w;                    pre[b] = a;                }            }            else if(vis[a] == 0)///如果两点在一棵树上 并且这棵树还没有环  那么就添加一个环            {                sum += edge[i].w;                vis[a] = 1;            }        }        printf("%d\n",sum);    }    return 0;}