Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem F: Heavy Cycle Edges

Given an undirected graph with edge weights, a minimum spanning tree is a subset of edges of minimum total weight such that any two nodes are connected by some path containing only these edges. A popular algorithm for finding the minimum spanning tree T in a graph proceeds as follows:

• let T be initially empty
• consider the edges e₁, ..., e_m in increasing order of weight
  • add e_i to T if the endpoints of e_i are not connected by a path in T

An alternative algorithm is the following:

• let T be initially the set of all edges
• while there is some cycle C in T
  • remove edge e from T where e has the heaviest weight in C

Your task is to implement a function related to this algorithm. Given an undirected graph G with edge weights, your task is to output all edges that are the heaviest edge in some cycle of G.

Input Format

The first input of each case begins with integers n and m with 1 ≤ n ≤ 1,000 and 0 ≤ m ≤ 25,000 where n is the number of nodes and m is the number of edges in the graph. Following this are m lines containing three integers u, v, and w describing a weight w edge connecting nodes u and v where 0 ≤ u, v < n and 0 ≤ w < 2³¹. Input is terminated with a line containing n = m = 0; this case should not be processed. You may assume no two edges have the same weight and no two nodes are directly connected by more than one edge.

Output Format

Output for an input case consists of a single line containing the weights of all edges that are the heaviest edge in some cycle of the input graph. These weights should appear in increasing order and consecutive weights should be separated by a space. If there are no cycles in the graph then output the text forest instead of numbers.

Sample Input

3 3
0 1 1
1 2 2
2 0 3
4 5
0 1 1
1 2 2
2 3 3
3 1 4
0 2 0
3 1
0 1 1
0 0

Sample Output

3
2 4
forest

---

原本 MST 通常是 P法跟 K法這兩種方法去求，
這裡他給了一個特殊的解法，每次找尋環的最大權值並且將它刪除。
這題我沒有什麼想法，先將最大權值邊做排序後，依序窮舉是否是環上的邊，若是則將它刪除。

#include <stdio.h>
#include <stdlib.h>
#include <vector>
#include <queue>
using namespace std;
typedef struct {
    int i, j, v;
} E;
E D[25005];
vector<int> g[1005];
int cmp(const void *i, const void *j) {
    static E *a, *b;
    a = (E *)i, b = (E *)j;
    return b->v - a->v;
}
int main() {
    int n, m, i, j;
    int x, y, v;
    while(scanf("%d %d", &n, &m) == 2) {
        if(n == 0)  break;
        for(i = 0; i < n; i++)
            g[i].clear();
        for(i = 0; i < m; i++) {
            scanf("%d %d %d", &D[i].i, &D[i].j, &D[i].v);
            g[D[i].i].push_back(D[i].j);
            g[D[i].j].push_back(D[i].i);
        }
        qsort(D, m, sizeof(E), cmp);
        int tn;
        vector<int> buf;
        char used[n];
        for(i = 0; i < m; i++) {
            for(j = 0; j < n; j++)  used[j] = 0;
            used[D[i].i] = 1;
            queue<int> Q;
            Q.push(D[i].i);
            while(!Q.empty()) {
                tn = Q.front();
                Q.pop();
                for(j = 0; j < g[tn].size(); j++) {
                    if(tn == D[i].i && g[tn][j] == D[i].j)  continue;
                    if(!used[g[tn][j]]) {
                        used[g[tn][j]] = 1;
                        Q.push(g[tn][j]);
                    }
                }
            }
            if(used[D[i].j]) {
                for(vector<int>::iterator it = g[D[i].i].begin();
                    it != g[D[i].i].end(); ) {
                    if(*it == D[i].j) {
                        it = g[D[i].i].erase(it);
                        break;
                    } else {
                        it++;
                    }
                }
                for(vector<int>::iterator it = g[D[i].j].begin();
                    it != g[D[i].j].end(); ) {
                    if(*it == D[i].i) {
                        it = g[D[i].j].erase(it);
                        break;
                    } else {
                        it++;
                    }
                }
                buf.push_back(D[i].v);
            }
        }
        if(buf.size()) {
            for(i = buf.size()-1; i >= 0; i--) {
                printf("%d", buf[i]);
                if(i)  putchar(' ');
            }
            puts("");
        } else
            puts("forest");
    }
    return 0;
}