HDU3367 Pseudoforest 【并查集】+【贪心】

Pseudoforest

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1729    Accepted Submission(s): 661

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

 

Source

“光庭杯”第五届华中北区程序设计邀请赛 暨 WHU第八届程序设计竞赛 

题意：这题的题意折腾了好久才懂，開始以为是求最多有一个环的最大联通分量，然后WA了，后来才知道是求多个分量的最大和。每一个分量最多有一个环。

题解：边权从大到小排序，对于每条边的起点终点a、b有三种情况：

1、a、b分属两个环。此时a、b不能连接；

2、a、b当中一个属于环。此时将还有一个连过去或者两个都不在环中，此时随意连；

3、a、b在同一集合。但该集合无环。此时连接a、b并生成环。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 10002
#define maxm 100002
using std::sort;

struct Node{
	int u, v, cost;
} E[maxm];
int pre[maxn];
bool Ring[maxn];

bool cmp(Node a, Node b){
	return a.cost > b.cost;
}

int ufind(int k)
{
	int a = k, b;
	while(pre[k] != -1) k = pre[k];
	while(a != k){
		b = pre[a];
		pre[a] = k;
		a = b;
	}
	return k;
}

int greedy(int n, int m)
{
	int ans = 0, i, u, v;
	for(i = 0; i < m; ++i){
		u = E[i].u; v = E[i].v;
		u = ufind(u); v = ufind(v);
		if(Ring[u] && Ring[v]) continue;
		if(u != v){
			if(Ring[v]) pre[u] = v;
			else pre[v] = u;
			ans += E[i].cost;
		}else if(Ring[v] == false){
			Ring[v] = true;
			ans += E[i].cost;
		}
	}
	return ans;
}

int main()
{
	int n, m, a, b, c, i;
	while(scanf("%d%d", &n, &m), n||m){
		memset(pre, -1, sizeof(pre));
		memset(Ring, 0, sizeof(Ring));
		for(i = 0; i < m; ++i)
			scanf("%d%d%d", &E[i].u, &E[i].v, &E[i].cost);
		sort(E, E + m, cmp);
		printf("%d
", greedy(n, m));
	}
	return 0;
}