hdu 3367 Pseudoforest(最大生成树)

最新推荐文章于 2021-07-28 20:12:48 发布

清，纯一色

最新推荐文章于 2021-07-28 20:12:48 发布

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文章标签： java

原文链接：http://www.cnblogs.com/zyx1314/p/3877202.html

本文介绍了一种求解伪森林图的最大边权和问题的算法。伪森林图是一种特殊的图，其每个连通组件中至多包含一个环。文章通过示例详细解释了如何在确保连通组件最多只有一个环的前提下，选取边来获得最大的边权总和。

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Pseudoforest

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1705 Accepted Submission(s): 653

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

题意：n个点，给你n条边，要求你连一些边，是边权总和最大（整个图可以不连通，对于每个连通子图最多只能有一个环）

：：按照最大生成树的做法，只要在连接同一个连通子图的两个结点是判断该子图是否已有环，没有才可以连接。

在连接不同连通子图的两个结点是判断两个连通子图的环数之和是否小于2，满足条件就可连接。

view code#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
#define REP(i,n) for(int i=0; i<(n); i++)
const int N = 10010;
int n, m, fa[N], num[N];

struct edge
{
    int u, v, w;
    bool operator < (const edge &o) const {
        return w>o.w;
    }
}e[N*10];

int find(int x)
{
    return x==fa[x]?x:(fa[x]=find(fa[x]));
}

void solve()
{
    REP(i,n) fa[i] = i, num[i] = 0;
    REP(i,m) scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
    sort(e, e+m);
    int ans = 0;
    REP(i, m)
    {
        int u = find(e[i].u), v =find(e[i].v);
        if(u==v && num[v]==0)
            ans += e[i].w, num[v]++;
        else if(u!=v && num[v]+num[u]<2)
            fa[u] = v, num[v]+= num[u], ans += e[i].w;
    }
    printf("%d\n", ans);
}

int main()
{
//    freopen("in.txt", "r", stdin);
    while(cin>>n>>m &&(n|m) ) solve();
    return 0;
}