codeforces 962 F Simple Cycles Edges

时间：2020-02-04 13:47:50 阅读：76 评论：0 收藏：0 [点我收藏+]

标签：sync als efi nsis pointer NPU sar padding incr

求简单环，即求点=边数的点双分量，加上判断点和边的模板即可

const int maxm = 1e5+5;

int head[maxm<<1], edgecnt, dfn[maxm], low[maxm], bcc_cnt, bccnum[maxm], dfs_clock, s[maxm], top;
vector<int> bcc[maxm], ans;

struct edge{
    int u, v, nex;
} edges[maxm<<1];

void addedge(int u, int v) {
    edges[++edgecnt].u = u;
    edges[edgecnt].v = v;
    edges[edgecnt].nex = head[u];
    head[u] = edgecnt;
}

void tarjan(int u, int fa) {
    dfn[u] = low[u] = ++dfs_clock;
    int child = 0, v;
    for(int i = head[u]; i; i = edges[i].nex) {
        v = edges[i].v;
        if(v == fa) continue;
        if(!dfn[v]) {
            s[++top] = i;
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if(low[v] >= dfn[u]) {
                bcc[++bcc_cnt].clear();
                int vet = 0;
                while(true) {
                    int num = s[top--];
                    bcc[bcc_cnt].push_back((num+1)/2);
                    if(bccnum[edges[num].u] != bcc_cnt) {
                        vet++;
                        bccnum[edges[num].u]=bcc_cnt;
                    }
                    if(bccnum[edges[num].v] != bcc_cnt) {
                        vet++;
                        bccnum[edges[num].v]=bcc_cnt;
                    }
                    if(edges[num].u == u && edges[num].v == v) break;
                }
                if(vet == bcc[bcc_cnt].size())
                    for(int i = 0; i < bcc[bcc_cnt].size(); ++i)
                        ans.push_back(bcc[bcc_cnt][i]);
            }
        } else if(dfn[v] < dfn[u]){
            s[++top] = i;
            low[u] = min(low[u], dfn[v]);
        }
    }
}

void run_case() {
    int n, m, u, v;
    cin >> n >> m;
    for(int i = 1; i <= m; ++i) {
        cin >> u >> v;
        addedge(u, v), addedge(v, u);
    }
    for(int i = 1; i <= n; ++i)
        if(!dfn[i])
            tarjan(i, -1);
    sort(ans.begin(), ans.end());
    cout << ans.size() << "\n";
    for(int i : ans)
        cout << i << " ";
    
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    run_case();
    return 0;
}

View Code

You are given an undirected graph, consisting of $n$ vertices and $m$ edges. The graph does not necessarily connected. Guaranteed, that the graph does not contain multiple edges (more than one edges between a pair of vertices) or loops (edges from a vertex to itself).

A cycle in a graph is called a simple, if it contains each own vertex exactly once. So simple cycle doesn‘t allow to visit a vertex more than once in a cycle.

Determine the edges, which belong to exactly on one simple cycle.

Input

The first line contain two integers $n$ and $m$ $(1 \leq n \leq 100 000$ , $0 \leq m \leq min (n \cdot (n - 1) / 2, 100 000))$ — the number of vertices and the number of edges.

Each of the following $m$ lines contain two integers $u$ and $v$ ( $1 \leq u, v \leq n$ , $u \neq v$ ) — the description of the edges.

Output

In the first line print the number of edges, which belong to exactly one simple cycle.

In the second line print the indices of edges, which belong to exactly one simple cycle, in increasing order. The edges are numbered from one in the same order as they are given in the input.

Examples
input
Copy
3 3
1 2
2 3
3 1
output
Copy
3
1 2 3 
input
Copy
6 7
2 3
3 4
4 2
1 2
1 5
5 6
6 1
output
Copy
6
1 2 3 5 6 7 
input
Copy
5 6
1 2
2 3
2 4
4 3
2 5
5 3
output
Copy
0

codeforces 962 F Simple Cycles Edges

标签：sync als efi nsis pointer NPU sar padding incr

原文地址：https://www.cnblogs.com/GRedComeT/p/12259130.html

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