題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=3367

Pseudoforest

Time Limit: 10000/5000 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2870????Accepted Submission(s): 1126

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.

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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.

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Output Output the sum of the value of the edges of the maximum pesudoforest.

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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

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Sample Output 3 5

?題目大意：在一個無向圖中，給定一些邊的聯通情況以及邊的權值，求最大生成樹（最多存在一條環路）。

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解題思路：用kruskal的方法按照求最大生成樹那樣求的，只不過要加一個判斷，就是判斷兩顆子樹是夠成環，

　　　　　如果各成環，就不能合并，如果只有其中一個成環或者都不成環，那么就可以合并，并對其進行標記。。。

AC代碼：

20041234 2017-03-08 16:17:45 Accepted 3367 546MS 2668K 1272 B G++#include <stdio.h> #include <string.h> #include <algorithm>using namespace std;struct point {int u,v,l; }p[100010]; int parent[10010],n,m,vis[10010]; // vis數組用來標記是否形成環 bool cmp(point a, point b) {return a.l > b.l; // 從大到小排列 }int find (int x) {int s,tmp;for (s = x; parent[s] >= 0; s = parent[s]);while (s != x){tmp = parent[x];parent[x] = s;x = tmp;}return s; } void Union(int A, int B) {int a = find(A), b = find(B);int tmp = parent[a]+parent[b];if (parent[a] < parent[b]){parent[b] = a;parent[a] = tmp;}else{parent[a] = b;parent[b] = tmp;} } int kruskal() {int sum = 0,max = 0;sort(p,p+m,cmp);memset(vis,0,sizeof(vis));memset(parent,-1,sizeof(parent));for (int i = 0; i < m; i ++){int u = find(p[i].u), v = find(p[i].v);if (u != v){if (vis[u] && vis[v]) continue; // 如果兩棵子樹，各自能夠形成一個環，則不合并 if (vis[u] || vis[v]) // 如果只有其中一個形成環，或者兩個都沒形成環，合并同時標記 vis[u] = vis[v] = 1;max += p[i].l;Union(u,v);}else if(!vis[u] || !vis[v]) // 在同一連通分量內且有一個或者兩個都沒形成環 合并且標記 {vis[u] = vis[v] = 1;max += p[i].l;Union(u,v);}}return max; } int main () {while (scanf("%d%d",&n,&m),n+m!=0){for (int i = 0; i < m; i ++)scanf("%d%d%d",&p[i].u,&p[i].v,&p[i].l);printf("%d\n",kruskal());}return 0; }

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轉載于:https://www.cnblogs.com/yoke/p/6520080.html

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