HDU3367 Pseudoforest 【並查集】+【貪心】

標籤：hdu3367

PseudoforestTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1729    Accepted Submission(s): 661

Problem DescriptionIn 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.

 
InputThe 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.
 
OutputOutput 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“光庭杯”第五屆華中北區程式設計邀請賽 暨 WHU第八屆程式設計競賽 

題意：這題的題意折騰了好久才懂，開始以為是求最多有一個環的最大聯通分量，然後WA了，後來才知道是求多個分量的最大和，每個分量最多有一個環。

題解：邊權從大到小排序，對於每條邊的起點終點a、b有三種情況：

1、a、b分屬兩個環，此時a、b不能串連；

2、a、b其中一個屬於環，此時將另一個連過去或者兩個都不在環中，此時任意連；

3、a、b在同一集合，但該集合無環，此時串連a、b並產生環。

#include <stdio.h>#include <string.h>#include <algorithm>#define maxn 10002#define maxm 100002using 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\n", greedy(n, m));}return 0;}