UVA 12544 - Beehives O(nm) 无向图最小环

Bees are one of the most industrious insects. Since they collect nectarand pollen from flowers, they
have to rely on the trees in the forest. For simplicity they numbered the n trees from 0 to n − 1. Instead
of roaming around all over the forest, they use a particular list of paths. A path is based on two trees,
and they can move either way i.e. from one tree to another in straight line. They don’t use paths that
are not in their list.
As technology has been improved a lot, they also changed their working strategy. Instead of hovering
over all the trees in the forest, they are targeting particular trees, mainly trees with lots of flowers.
So, they planned that they will build some new hives in some targeted trees. After that they will only
collect their foods from these trees. They will also remove some paths from their list so that they don’t
have to go to a tree with no hive in it.
Now, they want to build the hives such that if one of the paths in their new list go down (some
birds or animals disturbs them in that path) it’s still possible to go from any hive to another using the
existing paths.
They don’t want to choose less than two trees and as hive-building requires a lot of work, they need
to keep the number of hives as low as possible. Now you are given the trees with the paths they use,
your task is to propose a new bee hive colony for them.
Input
Input starts with an integer T (T ≤ 50), denoting the number of test cases.
Each case starts with a blank line. Next line contains two integers n (2 ≤ n ≤ 500) and m
(0 ≤ m ≤ 20000), where n denotes the number of trees and m denotes the number of paths. Each of
the next m lines contains two integers u v (0 ≤ u, v < n, u ̸= v) meaning that there is a path between
tree u and v. Assume that there can be at most one path between tree u to v, and needless to say that
a path will not be given more than once in the input.
Output
For each case, print the case number and the number of beehives in the proposed colony or ‘impossible’
if its impossible to find such a colony.
NOTE: Dataset is huge. Use faster I/O methods.
Sample Input
3
3
0
1
2
3
1
2
0
2 1
0 1
5
0
1
1
2
0
3
6
1
2
3
3
4
4
Sample Output
Case 1: 3
Case 2: impossible
Case 3: 3

#include <cstdio>
#include <cstring>
#include <cstring>
#include <queue>
#include <algorithm>
#include <vector>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 505;
const int M = 20005;


vector<int> g[N];
int vis[N], dis[N], pre[N];
queue<int> que;

int ans;
void bfs(int st) {
    while(!que.empty()) que.pop();
    memset(vis, 0, sizeof vis);
    memset(dis, INF, sizeof dis);
    memset(pre, -1, sizeof pre);
    dis[st] = 0;
    vis[st] = 1;
    que.push(st);

    while(!que.empty()) {
        int u = que.front(); que.pop();
        int sx = g[u].size();
        for(int i = 0; i < sx; ++i) {
            int v = g[u][i];
            if(v == pre[u]) continue;
            if(!vis[v]) {
                vis[v] = 1;
                dis[v] = dis[u] + 1;
                que.push(v);
                pre[v] = u;
            }else {
                ans = min(ans, dis[u] + dis[v] + 1);
            }
        }
    }
}
int main() {
//freopen("in", "r", stdin);
    int _, cas = 1; scanf("%d", &_);
    while(_ --) {
        int n, m; scanf("%d%d", &n, &m);
        int u, v;
        for(int i = 0; i <= n; ++i) g[i].clear();
        for(int i = 0; i < m; ++i) {
            scanf("%d%d", &u, &v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        ans = INF;
        for(int i = 0; i < n; ++i) {
            bfs(i);
        }
        printf("Case %d: ", cas++);
        if(ans == INF) puts("impossible");
        else printf("%d\n", ans);
    }
    return 0;
}

 

题意:给出一个无向图,n<=500&&m<=20000， 求一个最小环
思路：枚举起点s，bfs出从s到每个点的距离dis, 对于当前边u，v，如果v被访问过了，且上次v被访问不是通过u，即v ！= pre[u]，那么res = dis[u] + dis[v] + 1

但是注意，枚举的点不一定在环里面，且真正形成的环的长度也是小于等于res的，比如4个点，4条边， 0-1，1-2，1-3，2-3，当从0点bfs时，得到的备选res=5，但环的长度是3，
因为我们是枚举每一个点，最终一定能得到最小环